Resistors in Series and Parallel Questions in English

(N/A) Three cells of potential $2\, V$ each,connected in series,are equivalent to a battery of potential $2\, V + 2\, V + 2\, V = 6\, V$.
The following circuit diagram shows three resistors of resistances $5 \,\Omega$,$8 \,\Omega$,and $12 \,\Omega$ respectively,connected in series with a battery of potential $6\, V$ and a closed plug key.

(N/A) To measure the current flowing through the resistors,an ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the $12\,\Omega$ resistor,a voltmeter should be connected in parallel to this resistor.
The total voltage of the battery is $V = 3 \times 2\, V = 6\, V$.
The resistances are connected in series,so the total resistance $R$ is:
$R = 5\,\Omega + 8\,\Omega + 12\,\Omega = 25\,\Omega$.
According to Ohm's law,$V = IR$,the current $I$ flowing through the circuit is:
$I = \frac{V}{R} = \frac{6\, V}{25\,\Omega} = 0.24\, A$.
Thus,the ammeter reading is $0.24\, A$.
The potential difference $V_1$ across the $12\,\Omega$ resistor is calculated using Ohm's law:
$V_1 = I \times R_{12\,\Omega} = 0.24\, A \times 12\,\Omega = 2.88\, V$.
Therefore,the reading of the voltmeter is $2.88\, V$.

(A) When two resistors $R_1$ and $R_2$ are connected in parallel,the equivalent resistance $R$ is given by the formula: $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$.
Given $R_1 = 1\, \Omega$ and $R_2 = 10^6\, \Omega$.
Substituting the values: $\frac{1}{R} = \frac{1}{1} + \frac{1}{10^6} = 1 + 0.000001 = 1.000001\, \Omega^{-1}$.
Therefore,$R = \frac{1}{1.000001} \approx 0.999999\, \Omega$.
Since $0.999999\, \Omega < 1\, \Omega$,the equivalent resistance is slightly less than $1\, \Omega$.

(D) When $1 \, \Omega$,$10^3 \, \Omega$,and $10^6 \, \Omega$ are connected in parallel,the equivalent resistance $R$ is given by the formula:
$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
Substituting the values:
$\frac{1}{R} = \frac{1}{1} + \frac{1}{10^3} + \frac{1}{10^6} = 1 + 0.001 + 0.000001 = 1.001001 \, \Omega^{-1}$
Therefore,$R = \frac{1}{1.001001} \approx 0.999 \, \Omega$.
In a parallel combination,the equivalent resistance is always less than the smallest individual resistance. Since $1 \, \Omega$ is the smallest,the result $0.999 \, \Omega$ is consistent with this principle.

(D) Resistance of electric lamp,$R_1 = 100 \,\Omega$.
Resistance of toaster,$R_2 = 50 \,\Omega$.
Resistance of water filter,$R_3 = 500 \,\Omega$.
Voltage of the source,$V = 220 \,V$.
These are connected in parallel.
Let $R$ be the equivalent resistance of the circuit.
$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{100} + \frac{1}{50} + \frac{1}{500}$.
$\frac{1}{R} = \frac{5 + 10 + 1}{500} = \frac{16}{500}$.
$R = \frac{500}{16} \,\Omega = 31.25 \,\Omega$.
According to Ohm's law,$V = IR$,so the total current $I = \frac{V}{R}$.
$I = \frac{220}{500/16} = \frac{220 \times 16}{500} = 7.04 \,A$.
Since the electric iron takes as much current as all three appliances,the current through the iron is $I' = 7.04 \,A$.
Using Ohm's law for the electric iron,$R' = \frac{V}{I'} = \frac{220}{7.04} = 31.25 \,\Omega$.
Therefore,the resistance of the electric iron is $31.25 \,\Omega$ and the current flowing through it is $7.04 \,A$.

(N/A) $1$. In a parallel circuit,the voltage remains the same across all electrical appliances,which is equal to the source voltage. This ensures that each device operates at its rated voltage.
$2$. If one electrical appliance stops working due to a defect or switch-off,the other appliances continue to function because each device has its own independent circuit path.
$3$. The total effective resistance of the circuit decreases when appliances are connected in parallel,allowing for a higher total current to be drawn from the source as needed.

(N/A) We have three resistors with resistances $R_1 = 2 \,\Omega$,$R_2 = 3 \,\Omega$,and $R_3 = 6 \,\Omega$.
To obtain a total resistance of $4 \,\Omega$,we connect the $3 \,\Omega$ and $6 \,\Omega$ resistors in parallel,and then connect this combination in series with the $2 \,\Omega$ resistor.
First,calculate the equivalent resistance $(R_p)$ of the $3 \,\Omega$ and $6 \,\Omega$ resistors connected in parallel:
$\frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$
$R_p = 2 \,\Omega$
Next,connect this equivalent resistance $(R_p)$ in series with the $2 \,\Omega$ resistor:
$R_{total} = R_p + 2 \,\Omega = 2 \,\Omega + 2 \,\Omega = 4 \,\Omega$
Thus,the total resistance of the circuit is $4 \,\Omega$.

(N/A) To obtain a total resistance of $1\, \Omega$ from resistors of $2\, \Omega$,$3\, \Omega$,and $6\, \Omega$,they must be connected in parallel.
The formula for the equivalent resistance $(R_p)$ of resistors connected in parallel is given by:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
Substituting the given values:
$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$
Taking the least common multiple $(LCM)$ of $2, 3,$ and $6$,which is $6$:
$\frac{1}{R_p} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1\, \Omega^{-1}$
Therefore,$R_p = 1\, \Omega$.
Thus,connecting all three resistors in parallel results in a total resistance of $1\, \Omega$.

(A) Given four coils with resistances $R_1 = 4 \,\Omega, R_2 = 8 \,\Omega, R_3 = 12 \,\Omega$,and $R_4 = 24 \,\Omega$.
$(a)$ To obtain the highest total resistance,the coils must be connected in series. The equivalent resistance $R_s$ is the sum of individual resistances:
$R_s = R_1 + R_2 + R_3 + R_4 = 4 + 8 + 12 + 24 = 48 \,\Omega$.
$(b)$ To obtain the lowest total resistance,the coils must be connected in parallel. The equivalent resistance $R_p$ is given by:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24}$.
Taking the least common multiple $(LCM)$ of $4, 8, 12, 24$,which is $24$:
$\frac{1}{R_p} = \frac{6 + 3 + 2 + 1}{24} = \frac{12}{24} = \frac{1}{2}$.
Therefore,$R_p = 2 \,\Omega$.
Thus,the highest resistance is $48 \,\Omega$ and the lowest resistance is $2 \,\Omega$.

(C) The resistance of a wire is directly proportional to its length $(R \propto l)$.
When a wire of resistance $R$ is cut into five equal parts,the resistance of each part becomes $R/5$.
When these five parts are connected in parallel,the equivalent resistance $R'$ is given by the formula:
$1/R' = 1/R_1 + 1/R_2 + 1/R_3 + 1/R_4 + 1/R_5$
Substituting $R_1 = R_2 = R_3 = R_4 = R_5 = R/5$:
$1/R' = 5/R + 5/R + 5/R + 5/R + 5/R = 25/R$
Therefore,$R/R' = 25$.

(C) In a series circuit,the current remains the same through all components. According to Ohm's law,the current $I$ is given by $I = \frac{V}{R_{eq}}$.
The equivalent resistance $R_{eq}$ for resistors connected in series is the sum of individual resistances:
$R_{eq} = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 \, \Omega$.
Given the potential difference $V = 9 \, V$,the current is:
$I = \frac{9}{13.4} \approx 0.671 \, A$.
Thus,the current flowing through the $12 \, \Omega$ resistor is approximately $0.67 \, A$.

(D) According to Ohm's law,the equivalent resistance $R$ required for the circuit is given by $V = I R$,which implies $R = V / I$.
Given supply voltage $V = 220 \,V$ and current $I = 5 \,A$,the required equivalent resistance is $R = 220 / 5 = 44 \,\Omega$.
When $x$ resistors of resistance $r = 176 \,\Omega$ are connected in parallel,the equivalent resistance $R$ is given by $1 / R = x / r$.
Substituting the values,we get $1 / 44 = x / 176$.
Solving for $x$,we get $x = 176 / 44 = 4$.
Therefore,$4$ resistors of $176 \,\Omega$ are required.

(N/A) To obtain the desired equivalent resistance,we analyze different combinations of the three $6 \, \Omega$ resistors.
$(i)$ To get $9 \, \Omega$:
We connect two $6 \, \Omega$ resistors in parallel and the third $6 \, \Omega$ resistor in series with this combination.
Equivalent resistance of the parallel part: $\frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \implies R_p = 3 \, \Omega$.
Total resistance: $R_{eq} = R_p + 6 \, \Omega = 3 \, \Omega + 6 \, \Omega = 9 \, \Omega$.
$(ii)$ To get $4 \, \Omega$:
We connect two $6 \, \Omega$ resistors in series and the third $6 \, \Omega$ resistor in parallel with this combination.
Equivalent resistance of the series part: $R_s = 6 \, \Omega + 6 \, \Omega = 12 \, \Omega$.
Total resistance: $\frac{1}{R_{eq}} = \frac{1}{R_s} + \frac{1}{6} = \frac{1}{12} + \frac{1}{6} = \frac{1+2}{12} = \frac{3}{12} = \frac{1}{4} \implies R_{eq} = 4 \, \Omega$.

(B) To obtain the maximum resistance from a given set of resistors,they must be connected in series.
In a series connection,the equivalent resistance $(R_{eq})$ is the sum of individual resistances.
Given that there are $n = 5$ resistors,each with a resistance of $R = 1/5 \ \Omega$.
The formula for equivalent resistance in series is $R_{eq} = R_1 + R_2 + R_3 + R_4 + R_5$.
Substituting the values: $R_{eq} = 1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 5 \times (1/5) = 1 \ \Omega$.
Therefore,the maximum resistance is $1 \ \Omega$.

(C) To obtain the minimum resistance from a given set of resistors,they must be connected in a parallel combination.
For $n$ resistors each of resistance $R$ connected in parallel,the equivalent resistance $R_{eq}$ is given by the formula:
$1/R_{eq} = 1/R_1 + 1/R_2 + 1/R_3 + 1/R_4 + 1/R_5$
Given $R = 1/5\,\Omega$ and $n = 5$:
$1/R_{eq} = 5 / (1/5) = 25$
Therefore,$R_{eq} = 1/25\,\Omega$.

(A) In a series circuit,the current flowing through all components is the same because there is only one path for the charge to flow.
Therefore,when two resistors of $2 \,\Omega$ and $4 \,\Omega$ are connected in series,the same current flows through both of them.
In a parallel circuit,the potential difference across each resistor is the same,but the current flowing through them depends on their individual resistance values $(I = V/R)$.
Thus,option $A$ is the correct statement.

(N/A) No, the bulbs will not glow with the same brightness.
$1$. In a series circuit, the total resistance is $R_{series} = R + R + R = 3R$. Since the voltage $V$ is constant, the current in each bulb is $I_{series} = V / (3R) = (1/3) \times (V/R)$.
$2$. In a parallel circuit, each bulb is connected directly to the source voltage $V$. Therefore, the current in each bulb is $I_{parallel} = V/R$.
$3$. Since the power dissipated by a bulb is $P = I^2R$, the power in the series circuit is $P_{series} = (V/3R)^2 \times R = V^2 / (9R)$, while in the parallel circuit, it is $P_{parallel} = (V/R)^2 \times R = V^2 / R$.
$4$. Because $P_{parallel} > P_{series}$, the bulbs in the parallel circuit will glow much more brightly than those in the series circuit.

(N/A) $1$. In the series circuit,the bulbs are connected in a single path. When one bulb fuses,the filament breaks,creating an open circuit. Consequently,the flow of current stops entirely,and all the remaining bulbs will stop glowing.
$2$. In the parallel circuit,each bulb is connected across the same potential difference independently. If one bulb fuses,it only affects its own branch. The other two bulbs remain connected to the source and will continue to glow with the same brightness as before.

(N/A) To demonstrate that the same current flows through every part of a series circuit,follow these steps:
$1$. Connect three resistors $(R_1, R_2, R_3)$ in series with a battery,an ammeter,and a plug key.
$2$. Place the ammeter at different positions in the circuit: first between the battery and the first resistor,then between the resistors,and finally after the last resistor.
$3$. Observe the reading of the ammeter in each position.
$4$. You will notice that the ammeter shows the same reading in all positions.
$5$. Conclusion: Since the ammeter measures the current flowing through the circuit,the identical readings confirm that the same amount of current flows through every component in a series circuit.

(B) To conclude that the same potential difference exists across three resistors connected in parallel,follow these steps:
$1$. Connect three resistors $R_1, R_2,$ and $R_3$ in parallel across a battery of voltage $V$.
$2$. Connect a voltmeter in parallel across each resistor individually.
$3$. Observe the reading of the voltmeter for each resistor.
$4$. You will find that the voltmeter reading is the same for all three resistors,which is equal to the voltage $V$ of the battery.
$5$. This confirms that in a parallel circuit,the potential difference across each branch remains constant.

(N/A) Equivalent resistance is defined as the single resistance value that,when connected in a circuit,would draw the same amount of current from the source as the original combination of multiple resistors connected in series,parallel,or a combination of both. It effectively replaces the entire network of resistors with one equivalent component while maintaining the same electrical characteristics for the rest of the circuit.

(B) Resistors are said to be connected in series when they are joined end-to-end such that the same amount of electric current flows through each resistor.
In a series circuit,there is only one path for the current to flow,which ensures that the current remains constant throughout the circuit.

(B) Resistors are said to be connected in parallel when the potential difference $(V)$ across each resistor is the same.
In a parallel circuit,the current $(I)$ divides among the branches,but the voltage remains constant across all components connected between the same two points.

(A) When a wire of resistance $10 \, \Omega$ is bent into a circle,the total resistance is distributed uniformly along the circumference.
Any diameter divides the circle into two equal semicircular parts.
Each semicircle will have a resistance of $R_1 = R_2 = 10 / 2 = 5 \, \Omega$.
These two semicircular parts are connected in parallel between the two ends of the diameter.
The effective resistance $R_{eq}$ is given by the formula for parallel resistors: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.
Substituting the values: $\frac{1}{R_{eq}} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$.
Therefore,$R_{eq} = \frac{5}{2} = 2.5 \, \Omega$.

(A) The total resistance of the wire is $R = 5 \, \Omega$.
When the wire is bent into a circle,the resistance is distributed uniformly along the circumference.
When we consider two points at the ends of a diameter,the circle is divided into two equal semicircular parts.
Each semicircular part has a resistance of $R' = R / 2 = 5 / 2 = 2.5 \, \Omega$.
These two semicircular parts are connected in parallel between the two points.
The equivalent resistance $R_{eq}$ is given by the formula for parallel resistors: $1 / R_{eq} = 1 / R' + 1 / R'$.
$1 / R_{eq} = 1 / 2.5 + 1 / 2.5 = 2 / 2.5$.
$R_{eq} = 2.5 / 2 = 1.25 \, \Omega$.

(N/A) To construct the circuit,connect the two $1.5\, V$ cells in series to form a battery of $3\, V$.
Connect this battery in series with the three resistors of $5\, \Omega$,$10\, \Omega$,and $15\, \Omega$ and a plug key.
The schematic diagram is shown below.

(A) To obtain the lowest equivalent resistance from a given set of resistors,they must be connected in a parallel combination.
For resistors connected in parallel,the equivalent resistance $R_p$ is given by the formula:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}$
Substituting the given values $R_1 = 4\, \Omega, R_2 = 8\, \Omega, R_3 = 12\, \Omega$,and $R_4 = 24\, \Omega$:
$\frac{1}{R_p} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24}$
The least common multiple $(LCM)$ of $4, 8, 12$,and $24$ is $24$.
$\frac{1}{R_p} = \frac{6 + 3 + 2 + 1}{24} = \frac{12}{24} = \frac{1}{2}$
Therefore,$R_p = 2\, \Omega$.

(N/A) To draw the circuit diagram in series, connect the components one after another in a single loop. The circuit includes a cell of $1.5 \, V$, two resistors of $10 \, \Omega$ and $15 \, \Omega$, and a plug key, all joined in a series combination as shown in the diagram.

(A) According to Ohm's law, $V = IR$, which implies $R = V/I$.
In a $V-I$ graph, the slope of the line represents the resistance $(R = \text{slope} = \tan \theta)$.
Graph $A$ has a smaller slope compared to graph $B$, which means graph $A$ represents a lower resistance.
In a parallel combination of resistors, the equivalent resistance is always less than the individual resistances, whereas in a series combination, the equivalent resistance is greater than the individual resistances.
Therefore, graph $A$ represents the parallel combination because it corresponds to the lower resistance value.

(A) The resistance of a bulb is given by the formula $R = \frac{V^2}{P}$.
Since all bulbs have the same voltage rating $(V = 220 \ V)$,the resistance is inversely proportional to the power $(R \propto \frac{1}{P})$.
Therefore,the $40 \ W$ bulb has the highest resistance.
When bulbs are connected in series,the same current $(I)$ flows through each bulb.
The power dissipated (brightness) in each bulb is given by $P_{actual} = I^2 R$.
Since $I$ is constant for all bulbs in series,the brightness is directly proportional to the resistance $(P_{actual} \propto R)$.
Because the $40 \ W$ bulb has the highest resistance,it will dissipate the most power and glow the brightest.

(N/A) $(i)$ The circuit diagram showing two resistors $R_1$ and $R_2$ connected in series is provided in the image.
$(ii)$ When resistors are connected in series,the same amount of electric current $(I)$ flows through each resistor. Therefore,the current passing through the $5 \ \Omega$ resistor is equal to the current passing through the $10 \ \Omega$ resistor.

(D) The arrangement shown in figure $(i)$ is a series arrangement. The arrangement in figure $(ii)$ is a parallel arrangement.
We would prefer the parallel arrangement for household electricity. The reasons are:
$(i)$ In a series arrangement,if one of the bulbs fuses,the circuit breaks and the other bulbs also cease to glow. However,in a parallel arrangement,if one bulb fuses,the other bulbs continue to glow because they are connected in independent branches.
$(ii)$ In a series arrangement,the total resistance of the circuit increases as more bulbs are added,which reduces the current and makes the bulbs glow less brightly. In a parallel arrangement,each bulb receives the full voltage of the source,so the brightness of one bulb remains unaffected when another bulb is switched on or off.

(A) In a parallel combination of resistors, the equivalent resistance is always less than the smallest individual resistance in the circuit.
Since the smallest resistor is $2 \, \Omega$, the total resistance must be less than $2 \, \Omega$.
This can be calculated using the formula for parallel resistors:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{5}$
$\frac{1}{R_p} = \frac{15 + 10 + 6}{30} = \frac{31}{30}$
$R_p = \frac{30}{31} \approx 0.967 \, \Omega$
Thus, the total resistance is approximately $0.967 \, \Omega$, which is less than $2 \, \Omega$.

(A) In a series circuit,the current $(I)$ flowing through all components is the same.
According to Ohm's Law,the potential difference $(V)$ across a resistor is given by $V = I \times R$.
Since $I$ is constant,$V$ is directly proportional to the resistance $(R)$: $V \propto R$.
The power rating of a bulb is given by $P = \frac{V^2}{R}$,which implies $R = \frac{V^2}{P}$.
For a fixed voltage rating,the resistance is inversely proportional to the power rating $(R \propto \frac{1}{P})$.
Therefore,the bulb with the lowest wattage $(40\, W)$ has the highest resistance,and the bulb with the highest wattage $(100\, W)$ has the lowest resistance.
Consequently,the potential difference is highest across the $40\, W$ bulb and lowest across the $100\, W$ bulb.

(N/A) $(i)$ Since the bulbs are connected in parallel,the potential difference across each bulb remains the same $(4.5 \, V)$. Therefore,if bulb $B_{1}$ fuses,the other two bulbs ($B_{2}$ and $B_{3}$) will continue to glow with the same intensity.
$(ii)$ Initially,the total current $I = 3 \, A$ is divided equally among the three identical bulbs in parallel,so each bulb draws $I_{1} = I_{2} = I_{3} = 1 \, A$.
When bulb $B_{2}$ fuses,it creates an open circuit in its branch,so the current through it becomes zero. Thus,the reading of $A_{2}$ becomes $0 \, A$.
Since $B_{1}$ and $B_{3}$ remain connected in parallel,they still experience the same potential difference and draw $1 \, A$ each. Therefore,$A_{1}$ shows $1 \, A$ and $A_{3}$ shows $1 \, A$.
The total current $A$ recorded by the main ammeter will be the sum of the currents in the remaining branches: $A = 1 \, A + 1 \, A = 2 \, A$.

(A) Ammeter $A_{1}$ will show a higher reading.
Resistance $R$ of a wire is given by the formula $R = \rho \frac{l}{A_{c}}$,where $\rho$ is resistivity,$l$ is length,and $A_{c}$ is the cross-sectional area.
Since wires $A$ and $B$ are of the same material and length,$\rho$ and $l$ are constant.
Therefore,$R \propto \frac{1}{A_{c}}$.
Since wire $A$ is thicker than wire $B$,it has a larger cross-sectional area $(A_{c})$,which means it has lower resistance.
According to Ohm's law,$I = \frac{V}{R}$. Since both wires are connected in parallel,the potential difference $V$ across them is the same.
Because wire $A$ has lower resistance,the current $I$ flowing through it will be higher.
Thus,ammeter $A_{1}$ will show a higher reading.

(C) We know that for a $V-I$ graph,the slope represents the resistance $(R = V/I)$.
In a series combination,the equivalent resistance $(R_s = R_1 + R_2)$ is always greater than the individual resistances ($R_1$ and $R_2$).
Therefore,the graph with the highest slope represents the series combination.
In the given graph,line $C$ has the maximum slope compared to lines $A$ and $B$.
Thus,$C$ represents the series combination of the other two resistors.

(A) In a series combination,all components are connected in a single path. If the filament of one lamp breaks,the circuit becomes open,and the flow of current is interrupted throughout the entire circuit. Therefore,the other two lamps will stop glowing.
$(b)$ In a parallel combination,each lamp is connected in a separate branch across the power source. If the filament of one lamp breaks,it only affects that specific branch. The other two lamps remain connected to the voltage source and will continue to glow normally.

$(7.5 \, \Omega)$ The circuit consists of resistors in series and parallel combinations.
First, consider the two $2 \, \Omega$ resistors connected in parallel. Let their combined resistance be $R_{P1}$.
$\frac{1}{R_{P1}} = \frac{1}{2} + \frac{1}{2} = 1 \implies R_{P1} = 1 \, \Omega$.
Next, consider the two $1 \, \Omega$ resistors connected in parallel. Let their combined resistance be $R_{P2}$.
$\frac{1}{R_{P2}} = \frac{1}{1} + \frac{1}{1} = 2 \implies R_{P2} = 0.5 \, \Omega$.
Finally, the total equivalent resistance $R$ of the circuit is the sum of the series components:
$R = 3 \, \Omega + 3 \, \Omega + R_{P1} + R_{P2}$
$R = 6 \, \Omega + 1 \, \Omega + 0.5 \, \Omega = 7.5 \, \Omega$.

(N/A) Consider three resistors of resistances $R_{1}, R_{2}$ and $R_{3}$ connected in series to a cell of potential $V$ as shown in the figure. Since the three resistors are connected in series,the current $I$ flowing through each of them is the same.
By Ohm's law,the potential drop across each resistor is given by:
$V_{1} = IR_{1}, V_{2} = IR_{2}$ and $V_{3} = IR_{3}$
Since $V$ is the total potential in the circuit,by the conservation of energy,we have:
$V = V_{1} + V_{2} + V_{3}$ $...(1)$
Substituting the values of $V_{1}, V_{2}$ and $V_{3}$ in equation $(1)$,we get:
$V = IR_{1} + IR_{2} + IR_{3}$ $...(2)$
If $R_{S}$ is the equivalent resistance of the series combination,then by Ohm's law:
$V = IR_{S}$ $...(3)$
Comparing equations $(2)$ and $(3)$,we have:
$IR_{S} = IR_{1} + IR_{2} + IR_{3}$
Dividing both sides by $I$,we get:
$R_{S} = R_{1} + R_{2} + R_{3}$ $...(4)$
Thus,in a series combination,the equivalent resistance is the sum of the individual resistances.

(N/A) Consider three resistors $R_{1}, R_{2}$ and $R_{3}$ connected in parallel as shown in the figure.
When the total current $I$ reaches point $a$,it splits into three parts: $I_{1}$ flowing through $R_{1}$,$I_{2}$ flowing through $R_{2}$,and $I_{3}$ flowing through $R_{3}$.
Since charge must be conserved,the total current $I$ entering point $a$ must be equal to the sum of the currents leaving that point. Therefore,we have:
$I = I_{1} + I_{2} + I_{3}$ $...(1)$
Since the resistors are connected in parallel,the potential difference $V$ across each resistor is the same. According to Ohm's law,the current through each resistor is:
$I_{1} = \frac{V}{R_{1}}, I_{2} = \frac{V}{R_{2}}, I_{3} = \frac{V}{R_{3}}$
Substituting these values into equation $(1)$,we get:
$I = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}$ $...(2)$
If $R_{P}$ is the equivalent resistance of the parallel combination,then the total current $I$ can be written as:
$I = \frac{V}{R_{P}}$ $...(3)$
Comparing equations $(2)$ and $(3)$,we have:
$\frac{V}{R_{P}} = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}$
Dividing both sides by $V$,we get the expression for equivalent resistance:
$\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$

(N/A) In a series combination,the total resistance is the sum of individual resistances:
$R_{S} = R + R + \dots + R \ (n \ \text{times}) = nR$
In a parallel combination,the reciprocal of the total resistance is the sum of the reciprocals of individual resistances:
$\frac{1}{R_{P}} = \frac{1}{R} + \frac{1}{R} + \dots + \frac{1}{R} \ (n \ \text{times}) = \frac{n}{R}$
Therefore,$R_{P} = \frac{R}{n}$
$(b)$ For resistors connected in parallel,the resultant resistance $R_{P}$ is given by:
$\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$
$\frac{1}{R_{P}} = \frac{1}{3} + \frac{1}{4} + \frac{1}{12}$
Taking the least common multiple $(LCM)$ of $3, 4$,and $12$,which is $12$:
$\frac{1}{R_{P}} = \frac{4 + 3 + 1}{12} = \frac{8}{12} = \frac{2}{3}$
$R_{P} = \frac{3}{2} = 1.5 \ \Omega$

(N/A) $(i)$ The two resistors are connected in a parallel combination because both ends of the resistors are connected to the same two points in the circuit.
$(ii)$ According to Ohm's law,$I = \frac{V}{R}$. Since the resistors are in parallel,the potential difference $(V)$ across each resistor is $3 \, V$.
$(a)$ Current through the $10 \, \Omega$ resistor: $I_1 = \frac{V}{R_1} = \frac{3 \, V}{10 \, \Omega} = 0.3 \, A$.
$(b)$ Current through the $15 \, \Omega$ resistor: $I_2 = \frac{V}{R_2} = \frac{3 \, V}{15 \, \Omega} = 0.2 \, A$.
$(iii)$ The ammeter measures the total current $(I)$ flowing through the circuit,which is the sum of the currents through the individual branches:
$I = I_1 + I_2 = 0.3 \, A + 0.2 \, A = 0.5 \, A$.
Therefore,the ammeter reading is $0.5 \, A$.

(N/A) The circuit diagram is as shown in the image.
Join three resistors of different values in series. Connect them with a battery,an ammeter,a rheostat,and a plug key,as shown in the figure. Close the plug key. Note the ammeter reading. Change the position of the ammeter to anywhere in between the resistors. Note the ammeter reading each time. We observe that the value of the current in the ammeter is the same,independent of its position in the electric circuit. It means that in a series combination of resistors,the current is the same in every part of the circuit or the same current flows through each resistor.
Precautions:
$(i)$ The current should not be passed for a long time to avoid heating.
$(ii)$ All the connections should be tight.

(N/A) The resistors $6 \, \Omega$ and $3 \, \Omega$ are connected in parallel.
Therefore,the equivalent resistance $R_P$ is given by:
$\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2}$
$\frac{1}{R_P} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$
Thus,$R_P = 2 \, \Omega$.
Now,the $2 \, \Omega$ resistor and the parallel combination $R_P$ are in series.
Therefore,the total equivalent resistance $R$ is:
$R = 2 \, \Omega + R_P = 2 \, \Omega + 2 \, \Omega = 4 \, \Omega$.
The equivalent resistance of the circuit is $4 \, \Omega$.

$(11 \Omega)$ The given circuit can be redrawn by identifying the nodes connected by wires. Points $A$ and $C$ are at the same potential, and points $E$ and $F$ are connected to $D$ via a wire, effectively making them parallel.
It is clear that the resistances $20 \Omega$ (between $C$ and $D$), $10 \Omega$ (between $D$ and $E$), and $20 \Omega$ (between $E$ and $F$) are in a parallel combination between the effective nodes.
If $R_{1}$ is the effective resistance of these three parallel resistors, then:
$\frac{1}{R_{1}} = \frac{1}{20} + \frac{1}{10} + \frac{1}{20}$
$\frac{1}{R_{1}} = \frac{1 + 2 + 1}{20} = \frac{4}{20} = \frac{1}{5}$
Therefore, $R_{1} = 5 \Omega$.
Now, the resistor $R_{1}$ and the $6 \Omega$ resistor are in series. Therefore, the equivalent resistance $R$ between $A$ and $B$ is:
$R = R_{1} + 6 \Omega$
$R = 5 \Omega + 6 \Omega = 11 \Omega$.

(A) When $R_{1}$ and $R_{2}$ are connected in parallel,the net resistance $(R_{p})$ is given by:
$\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
Therefore,we have:
$R_{p} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = 3 \, \Omega$ $...(1)$
When $R_{1}$ and $R_{2}$ are connected in series,the net resistance $(R_{s})$ is given by:
$R_{s} = R_{1} + R_{2} = 16 \, \Omega$ $...(2)$
Substituting equation $(2)$ into equation $(1)$:
$\frac{R_{1} R_{2}}{16} = 3 \implies R_{1} R_{2} = 48$ $...(3)$
From equation $(2)$,$R_{2} = 16 - R_{1}$. Substituting this into equation $(3)$:
$R_{1}(16 - R_{1}) = 48$
$16R_{1} - R_{1}^{2} = 48$
$R_{1}^{2} - 16R_{1} + 48 = 0$
$(R_{1} - 12)(R_{1} - 4) = 0$
Thus,$R_{1} = 12 \, \Omega$ or $R_{1} = 4 \, \Omega$.
If $R_{1} = 12 \, \Omega$,then $R_{2} = 4 \, \Omega$. If $R_{1} = 4 \, \Omega$,then $R_{2} = 12 \, \Omega$.
Therefore,the resistances are $4 \, \Omega$ and $12 \, \Omega$.

(N/A) $(i)$ Resistors of value $10 \ \Omega$ and $20 \ \Omega$ are connected in series,therefore,their equivalent resistance is
$R_{S} = 10 + 20 = 30 \ \Omega$
Similarly,resistors of value $5 \ \Omega$ and $25 \ \Omega$ are in series,therefore,their equivalent resistance is
$R_{S_{1}} = 5 + 25 = 30 \ \Omega$
Now,$R_{S}$ and $R_{S_{1}}$ are in parallel,therefore,we have
$R_{P} = \frac{R_{S} \times R_{S_{1}}}{R_{S} + R_{S_{1}}} = \frac{30 \times 30}{30 + 30} = \frac{900}{60} = 15 \ \Omega$
$(ii)$ The total current is given by Ohm's law:
$I = \frac{V}{R_{P}} = \frac{12}{15} = 0.8 \ A$

(B) Given: $R_{1} = 2 \Omega, R_{2} = 2 \Omega, V = 12 \text{ V}$.
$(i)$ When resistors are connected in series,the equivalent resistance is:
$R_{S} = R_{1} + R_{2} = 2 + 2 = 4 \Omega$.
Power consumed in series is:
$P_{S} = \frac{V^{2}}{R_{S}} = \frac{12^{2}}{4} = \frac{144}{4} = 36 \text{ W}$.
$(ii)$ When resistors are connected in parallel,the equivalent resistance is:
$R_{P} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{2 \times 2}{2 + 2} = \frac{4}{4} = 1 \Omega$.
Power consumed in parallel is:
$P_{P} = \frac{V^{2}}{R_{P}} = \frac{12^{2}}{1} = 144 \text{ W}$.
The ratio of power consumed in the two cases (parallel to series) is:
$\frac{P_{P}}{P_{S}} = \frac{144}{36} = 4:1$.

(N/A) Since the wire is divided into four equal parts,the new length of each part becomes one-fourth of the original length. Since resistance is directly proportional to length $(R \propto l)$,the resistance of each part becomes one-fourth of the original resistance,$i.e., R' = R/4$.
$(b)$ When these four parts are connected in parallel,the equivalent resistance $R_p$ is given by the formula:
$\frac{1}{R_p} = \frac{1}{R'} + \frac{1}{R'} + \frac{1}{R'} + \frac{1}{R'}$
Substituting $R' = R/4$:
$\frac{1}{R_p} = \frac{4}{R} + \frac{4}{R} + \frac{4}{R} + \frac{4}{R} = \frac{16}{R}$
Therefore,$R_p = R/16$.
Thus,the joint resistance is $1/16$ times the resistance of the original wire.