Mix Examples - Electricity Questions in English

(B) Heat produced in the circuit is inversely proportional to the resistance $R$ for a constant potential difference $V$,given by the formula $H = \frac{V^2}{R} t$.
Let $R$ be the resistance of each wire. If connected in series,the equivalent resistance is $R_S = R + R = 2R$.
If connected in parallel,the equivalent resistance is $R_P = \frac{R \times R}{R + R} = \frac{R}{2}$.
The ratio of heat produced in series $(H_S)$ to parallel $(H_P)$ is given by:
$\frac{H_S}{H_P} = \frac{V^2 / R_S \times t}{V^2 / R_P \times t} = \frac{R_P}{R_S}$.
Substituting the values of $R_S$ and $R_P$:
$\frac{H_S}{H_P} = \frac{R/2}{2R} = \frac{1}{4}$.
Thus,the ratio of heat produced is $1:4$.

(N/A) Supply voltage,$V = 220\, V$.
Resistance of each coil,$R = 24\,\Omega$.
$(i)$ When coils are used separately:
According to Ohm's law,$I = V / R$.
$I_1 = 220 / 24 = 9.17\, A$.
$(ii)$ When coils are connected in series:
Total resistance,$R_s = R + R = 24 + 24 = 48\,\Omega$.
$I_2 = V / R_s = 220 / 48 = 4.58\, A$.
$(iii)$ When coils are connected in parallel:
Total resistance,$1 / R_p = 1 / R + 1 / R = 1 / 24 + 1 / 24 = 2 / 24 = 1 / 12$.
So,$R_p = 12\,\Omega$.
$I_3 = V / R_p = 220 / 12 = 18.33\, A$.

(C) Case $(i)$: Potential difference,$V = 6 \,V$. The $1 \,\Omega$ and $2 \,\Omega$ resistors are connected in series. The equivalent resistance $R = 1 + 2 = 3 \,\Omega$. According to Ohm's law,$I = V/R = 6/3 = 2 \,A$. In a series circuit,the current remains the same through all components. Thus,the current through the $2 \,\Omega$ resistor is $2 \,A$. Power $P = I^2 R = (2)^2 \times 2 = 8 \,W$.
Case $(ii)$: Potential difference,$V = 4 \,V$. The $12 \,\Omega$ and $2 \,\Omega$ resistors are connected in parallel. In a parallel circuit,the voltage across each component is the same as the source voltage. Thus,the voltage across the $2 \,\Omega$ resistor is $4 \,V$. Power $P = V^2/R = (4)^2/2 = 16/2 = 8 \,W$.
Conclusion: The power used in the $2 \,\Omega$ resistor is $8 \,W$ in both cases.

(A) According to Ohm's Law,the current $I$ in a series circuit is given by $I = V/R$,where $V$ is the potential difference (voltage of the cell) and $R$ is the total resistance of the circuit.
In all three circuit diagrams $(i)$,$(ii)$,and $(iii)$,the components (cell,resistor,key,and ammeter) are connected in series.
Since the components are the same and they are connected in series,the total resistance $R$ and the total voltage $V$ remain constant in all three cases.
Therefore,the current $I$ flowing through the circuit will be the same in all three cases.

(D) The heat produced in a circuit is given by the formula $H = \frac{V^2}{R} \times t$,where $V$ is the potential difference,$R$ is the equivalent resistance,and $t$ is the time.
Since $V = 12\, V$ is constant for all cases,the heat produced $H$ is inversely proportional to the equivalent resistance $R$ $(H \propto \frac{1}{R})$.
Case $(i)$: $A$ single resistor of $2\, \Omega$. Equivalent resistance $R_1 = 2\, \Omega$.
Case $(ii)$: Two resistors of $2\, \Omega$ each connected in series. Equivalent resistance $R_2 = 2\, \Omega + 2\, \Omega = 4\, \Omega$.
Case $(iii)$: Two resistors of $2\, \Omega$ each connected in parallel. Equivalent resistance $R_3 = \frac{2 \times 2}{2 + 2} = \frac{4}{4} = 1\, \Omega$.
Comparing the resistances: $R_3 (1\, \Omega) < R_1 (2\, \Omega) < R_2 (4\, \Omega)$.
Since $H \propto \frac{1}{R}$,the heat produced will be maximum where the resistance is minimum.
Therefore,the heat produced is maximum in case $(iii)$ because it has the lowest equivalent resistance $(1\, \Omega)$.

(A) Yes,the potential difference will be the same.
First,calculate the equivalent resistance of the parallel combination of two $4\,\Omega$ resistors:
$\frac{1}{R_p} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$
So,$R_p = 2\,\Omega$.
Since the $2\,\Omega$ resistor is in series with the parallel combination (which also has an equivalent resistance of $2\,\Omega$),the same current $I$ flows through both. According to Ohm's Law $(V = IR)$,since the resistance values are equal $(2\,\Omega = 2\,\Omega)$ and the current $I$ is the same,the potential difference across both will be equal.

(A) Step $1$: Calculate the resistance of the lamp.
In a series circuit,total resistance $R_{total} = V / I = 10 \ V / 1 \ A = 10 \ \Omega$.
Since $R_{total} = R_{lamp} + R_{conductor}$,we have $10 \ \Omega = R_{lamp} + 5 \ \Omega$.
Therefore,$R_{lamp} = 5 \ \Omega$.
Step $2$: Analyze the parallel connection.
When a $10 \ \Omega$ resistor is connected in parallel to the series combination,the potential difference across the series combination remains $10 \ V$ because the battery is connected directly across the parallel branches.
Since the potential difference across the series branch (lamp + $5 \ \Omega$ conductor) remains $10 \ V$,the current flowing through the series branch remains $I = V / R_{total} = 10 \ V / 10 \ \Omega = 1 \ A$.
Consequently,there will be no change in the current flowing through the $5 \ \Omega$ conductor,and there will be no change in the potential difference across the lamp.

(N/A) $(i)$ Since the bulbs are connected in parallel,the potential difference across each bulb remains the same $(4.5\,V)$. Therefore,the glow of the bulbs $B_2$ and $B_3$ will remain unchanged when $B_1$ fuses.
$(ii)$ When $B_2$ fuses,the circuit branch containing $B_2$ becomes open. Thus,$A_2$ will show $0\,A$. The other two branches remain unaffected,so $A_1$ will show $1\,A$ and $A_3$ will show $1\,A$. The total current recorded by ammeter $A$ will be $1\,A + 0\,A + 1\,A = 2\,A$.
$(iii)$ The total power dissipated in the circuit is given by $P = V \times I$. Given $V = 4.5\,V$ and total current $I = 3\,A$,we have $P = 4.5\,V \times 3\,A = 13.5\,W$.

(N/A) The two $8\,\Omega$ resistors are in parallel. Their effective resistance $R_p$ is given by:
$R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4\,\Omega$.
$(b)$ The total resistance of the circuit $R_{eq} = R_{series} + R_p = 4\,\Omega + 4\,\Omega = 8\,\Omega$. The total current $I$ flowing through the circuit is $I = \frac{V}{R_{eq}} = \frac{8\,V}{8\,\Omega} = 1\,A$. Since the $4\,\Omega$ resistor is in series with the rest of the circuit,the current flowing through it is $1\,A$.
$(c)$ The potential difference $V$ across the $4\,\Omega$ resistor is $V = I \times R = 1\,A \times 4\,\Omega = 4\,V$.
$(d)$ The power dissipated $P$ in the $4\,\Omega$ resistor is $P = I^2 R = (1\,A)^2 \times 4\,\Omega = 4\,W$.
$(e)$ There is no difference in the ammeter readings. In a series circuit,the same amount of current flows through every component.

(N/A) In a series circuit,the total voltage is divided among the components. If one component fails or is switched off,the entire circuit breaks,and no current flows. Furthermore,each device receives a lower voltage than required,and the total resistance increases,which reduces the current. Therefore,series arrangement is not used in domestic circuits.
$(b)$ The resistance $(R)$ of a wire is inversely proportional to its area of cross-section $(A)$. This is expressed as $R \propto \frac{1}{A}$. As the area of cross-section increases,the resistance decreases,and vice versa.

(A) $(i)$ False: Current is defined as the rate of flow of charge,i.e.,$I = Q/t$. The statement incorrectly suggests $I = Q \times t$.
$(ii)$ True: $A$ cell converts chemical energy into electrical energy through internal chemical reactions,which drives the charge flow.
$(iii)$ True: For pure metals,the resistivity increases as temperature rises due to increased collisions between electrons and vibrating ions.
$(iv)$ False: Ohm's law states that the potential difference across a conductor is directly proportional to the current flowing through it $(V \propto I)$,provided temperature remains constant. It does not relate to power.
$(v)$ True: In a series circuit,there is only one path for current,whereas a parallel circuit provides multiple branches for current flow.
$(vi)$ False: Resistance in a wire is primarily due to the collisions of electrons with the atoms/ions of the conductor's lattice,not due to electron-electron repulsion.

(B) Alternating current $(AC)$ is defined as the current that changes its magnitude continuously and reverses its direction periodically over time.
In contrast,direct current $(DC)$ is the current that flows in a single constant direction and maintains a constant magnitude over time.
Therefore,the fundamental difference lies in the periodic reversal of direction and variation in magnitude characteristic of $AC$ compared to the steady nature of $DC$.

(B) The current whose direction reverses periodically is called alternating current $(AC)$,whereas direct current $(DC)$ flows in a constant direction.
The primary advantage of $AC$ is that it can be transmitted over long distances with minimal power loss compared to $DC$.
$(b)$ Given: Power $(P)$ = $2 \,kW = 2000 \,W$,Voltage $(V)$ = $220 \,V$.
Using the formula $P = VI$,the current $(I)$ is calculated as:
$I = \frac{P}{V} = \frac{2000}{220} \approx 9.09 \,A$.
Since the calculated current $(9.09 \,A)$ is less than the fuse rating of $10 \,A$,the fuse will withstand the current and will not blow off when the air conditioner is switched on.

(N/A) An ammeter is always connected in series with the circuit components. It measures the electric current flowing through the circuit.
$A$ voltmeter is always connected in parallel across the points between which the potential difference is to be measured. It measures the potential difference across a conductor.

(N/A) The total resistance of the circuit is $R = R_1 + R_2 + R_3 = 1 \Omega + 3 \Omega + 2 \Omega = 6 \Omega$.
The total current in the circuit is $I = \frac{V}{R} = \frac{3 \text{ V}}{6 \Omega} = 0.5 \text{ A}$.
When the voltmeter is connected between $B$ and $C$,it measures the potential difference across the $3 \Omega$ resistor.
The potential difference across the $3 \Omega$ resistor is $V_{BC} = I \times R_{BC} = 0.5 \text{ A} \times 3 \Omega = 1.5 \text{ V}$.
Initially,the voltmeter was connected across the $1 \Omega$ resistor,measuring $V_{AB} = I \times R_{AB} = 0.5 \text{ A} \times 1 \Omega = 0.5 \text{ V}$.
Therefore,the reading of the voltmeter will increase from $0.5 \text{ V}$ to $1.5 \text{ V}$.

(A) Separately: Current $I = \frac{V}{R} = \frac{220 \ V}{22 \ \Omega} = 10 \ A$.
$(b)$ In series: Total resistance $R_s = R_1 + R_2 = 22 \ \Omega + 22 \ \Omega = 44 \ \Omega$.
Therefore,current $I = \frac{V}{R_s} = \frac{220 \ V}{44 \ \Omega} = 5 \ A$.
$(c)$ In parallel: Total resistance $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{22} + \frac{1}{22} = \frac{2}{22} = \frac{1}{11} \ \Omega^{-1}$.
Therefore,$R_p = 11 \ \Omega$.
Thus,current $I = \frac{V}{R_p} = \frac{220 \ V}{11 \ \Omega} = 20 \ A$.

(N/A) $Ohm's$ law states that the current flowing through a conductor is directly proportional to the potential difference across its ends,provided the temperature and other physical conditions remain constant.
Mathematically: $V = IR$,where $V$ is potential difference,$I$ is current,and $R$ is resistance.
$(b)$ $(i)$ Variable resistance: Represented by a resistor symbol with an arrow or a sliding contact.
$(ii)$ Voltmeter: Represented by a circle with a '$V$' inside,having positive and negative terminals.
$(c)$ Given: Rated power $P = 100 \, W$,Rated voltage $V = 220 \, V$.
First,calculate the resistance of the bulb:
$R = \frac{V^2}{P} = \frac{(220)^2}{100} = \frac{48400}{100} = 484 \, \Omega$.
Now,calculate the power consumed at $V_1 = 110 \, V$:
$P_{new} = \frac{V_1^2}{R} = \frac{(110)^2}{484} = \frac{12100}{484} = 25 \, W$.
The power consumed will be $25 \, W$.

(N/A) The instrument used to measure electric current in a circuit is an ammeter.
The unit of electric current is the ampere $(A)$. One ampere is defined as the flow of $1$ coulomb of electric charge through a cross-section of a conductor in $1$ second.
$(b)$ $(i)$ The symbol represents a rheostat or variable resistance.
$(ii)$ The symbol represents a closed plug key or a closed switch.
$(c)$ The circuit diagram consists of a battery of four cells (total $6\, V$),a plug key (closed),an ammeter connected in series with the $0.5\, m$ nichrome wire $XY$,and a voltmeter connected in parallel across the wire $XY$.

(A) The instrument used to measure the potential difference between two points in a circuit is a Voltmeter.
The $SI$ unit of potential difference is the Volt $(V)$. It is defined as the amount of work done $(W)$ in moving a unit positive charge $(q)$ from one point to another,i.e.,$V = \frac{W}{q}$. Thus,$1 \text{ Volt} = 1 \text{ Joule} / 1 \text{ Coulomb}$.
The circuit symbols are:
$(i)$ Variable resistor: $A$ resistor symbol with an arrow pointing across it.
$(ii)$ Closed plug key: $A$ circle with a dot inside it.
$(b)$ $(i)$ Circuit $(I)$ has more resistance because the resistors are connected in series,and the equivalent resistance is the sum of individual resistances $(R_{eq} = R_{1} + R_{2} + R_{3})$.
$(ii)$ More current passes through circuit $(II)$ because the equivalent resistance in parallel is less than the individual resistances,leading to a higher total current from the source.
$(iii)$ In circuit $(II)$,the potential difference across each resistor is equal because they are connected in parallel.
$(iv)$ In circuit $(I)$,more heat is produced in $R_{1}$ because in a series circuit,the current $(I)$ is the same through all resistors,and heat produced $H = I^{2}Rt$. Since $R_{1}$ is the largest resistance,it will produce the most heat.

(N/A) The potential difference between two points is said to be $1$ volt if $1$ joule of work is done in moving a charge of $1$ coulomb from one point to another.
$(b)$ The heating element of an electrical heater is made of an alloy (like nichrome) which has high resistivity. Due to high resistivity, it produces a large amount of heat when current passes through it, causing it to glow. In contrast, the connecting cord is made of copper, which has very low resistivity, so it does not produce significant heat and does not glow.
$(c)$ $(i)$ Silver is a better conductor than copper because it has lower electrical resistivity ($1.60 \times 10^{-8} \, \Omega m$ for silver vs $1.62 \times 10^{-8} \, \Omega m$ for copper).
$(ii)$ Nichrome is advised for electrical heating devices because it has a very high resistivity compared to other materials listed, which allows it to generate more heat according to Joule's law of heating $(H = I^2Rt)$.

$(a)$ Total resistance $R = 5\, \Omega + 8\, \Omega + 12\, \Omega = 25\, \Omega$.
Total voltage $V = 5 \times 2\, V = 10\, V$.
Using Ohm's law, the current $I = \frac{V}{R} = \frac{10\, V}{25\, \Omega} = 0.4\, A$.
Therefore, the ammeter reading is $0.4\, A$.
$(b)$ The potential difference across the $12\, \Omega$ resistor is $V' = I \times R_{12} = 0.4\, A \times 12\, \Omega = 4.8\, V$.
Therefore, the voltmeter reading is $4.8\, V$.

(N/A) The chemical action within a cell generates a potential difference across its terminals. This potential difference drives and maintains the flow of electric current in the circuit.
$(b)$ $(i)$ The paths $xTy$ and $xZy$ are connected in parallel between points $x$ and $y$. The equivalent resistance $R_e$ is given by:
$\frac{1}{R_e} = \frac{1}{2} + \frac{1}{6} = \frac{3+1}{6} = \frac{4}{6} = \frac{2}{3} \, \Omega^{-1}$
$R_e = 1.5 \, \Omega$
$(ii)$ The total resistance of the circuit is the sum of the equivalent resistance $R_e$ and the series resistor of $1.5 \, \Omega$:
$R_{total} = 1.5 \, \Omega + 1.5 \, \Omega = 3 \, \Omega$
Using Ohm's law,the main current $I = \frac{V}{R_{total}} = \frac{6 \, V}{3 \, \Omega} = 2 \, A$
$(iii)$ The potential difference across the parallel combination of $xTy$ and $xZy$ is $V_{xy} = I \times R_e = 2 \, A \times 1.5 \, \Omega = 3 \, V$.
Current through path $xTy$ $(2 \, \Omega)$: $I_1 = \frac{V_{xy}}{R_1} = \frac{3 \, V}{2 \, \Omega} = 1.5 \, A$
Current through path $xZy$ $(6 \, \Omega)$: $I_2 = \frac{V_{xy}}{R_2} = \frac{3 \, V}{6 \, \Omega} = 0.5 \, A$

(A) $(i)$ It means that the work done in moving a charge of $1 \, C$ between the two points is $1 \, J$.
$(ii)$ Using the formula $W = qV$,where $q = 5 \, C$ and $V = 12 \, V$,we get $W = 5 \times 12 = 60 \, J$.
$(b)$ To study the factors affecting resistance:
$(i)$ Set up a circuit with a cell,an ammeter,a nichrome wire of length $l$,and a plug key.
$(ii)$ Close the key and note the current reading in the ammeter.
$(iii)$ Replace the wire with another nichrome wire of the same thickness but double the length $(2l)$. Note the current.
$(iv)$ Replace the wire with a thicker nichrome wire of the same length $l$. Note the current.
$(v)$ Replace the wire with a copper wire of the same length and thickness as the first nichrome wire. Note the current.
$(vi)$ Observe that the current varies in each case,demonstrating that resistance depends on length,cross-sectional area,and the material of the conductor.

(N/A) $(i)$ Variable resistor or rheostat: It is used to change the resistance in a circuit,thereby controlling the current flowing through it.
$(ii)$ Wires crossing without joining: It represents two wires that cross each other without making any electrical contact.
$(b)$ The circuit diagram consists of a $12 \, V$ battery,a switch $(K)$,an ammeter $(A)$ in series with the combination,and a voltmeter $(V)$ connected in parallel across the three resistors ($5 \, \Omega$,$10 \, \Omega$,and $20 \, \Omega$) which are themselves connected in parallel.
$(c)$ Advantage of parallel connection: In a parallel circuit,if one electrical device stops working due to a defect or switch-off,the other devices continue to operate independently because each device has its own separate circuit path.

(N/A) The resistance $R$ is given by Ohm's law: $R = V / I$.
From the graph,at $V = 2 \, V$,the current $I = 0.1 \, A$.
Therefore,$R = 2 \, V / 0.1 \, A = 20 \, \Omega$.
$(b)$ Given: Total current $I = 5 \, A$,Voltage $V = 220 \, V$.
The required total resistance $R_{eq}$ of the circuit is $R_{eq} = V / I = 220 \, V / 5 \, A = 44 \, \Omega$.
Let $n$ resistors of $176 \, \Omega$ each be connected in parallel.
The equivalent resistance for $n$ resistors in parallel is $1 / R_{eq} = n / R$.
$1 / 44 = n / 176$.
$n = 176 / 44 = 4$.
Thus,$4$ resistors are required.
$(c)$ Electric power is defined as the rate at which electric energy is dissipated or consumed in an electric circuit.
We know that power $P = W / t$ and work done $W = V \cdot I \cdot t$.
So,$P = (V \cdot I \cdot t) / t = V \cdot I$.
From Ohm's law,$I = V / R$.
Substituting $I$ in the power equation: $P = V \cdot (V / R) = V^2 / R$.

(N/A) The circuit diagram shows three resistors $R_{1}, R_{2}, R_{3}$ connected in parallel across a potential difference $V$. The total current $I$ splits into $I_{1}, I_{2}, I_{3}$ through the resistors.
Since they are in parallel,the potential difference $V$ across each resistor is the same.
Total current $I = I_{1} + I_{2} + I_{3}$.
Using Ohm's law,$I_{1} = V/R_{1}$,$I_{2} = V/R_{2}$,and $I_{3} = V/R_{3}$.
If $R_{p}$ is the equivalent resistance,$I = V/R_{p}$.
Substituting these,$V/R_{p} = V/R_{1} + V/R_{2} + V/R_{3}$.
Thus,$1/R_{p} = 1/R_{1} + 1/R_{2} + 1/R_{3}$.
$(b)$ Electric bulbs are filled with chemically inactive gases like nitrogen or argon to prevent the oxidation of the tungsten filament at high temperatures,thereby prolonging its life.
$(c)$ The statement means that the fuse is designed to allow a maximum current of $5 \ A$ to flow through the circuit. If the current exceeds this value,the fuse wire melts due to excessive heating and breaks the circuit,protecting the appliances.

(N/A) $(i)$ The ammeter reading will decrease (become half). This is because resistance $R$ is directly proportional to length $L$ $(R \propto L)$. When length is doubled, resistance doubles, and according to Ohm's law $(I = V/R)$, the current $I$ becomes half.
$(ii)$ The ammeter reading will increase (become two times). This is because resistance $R$ is inversely proportional to the area of cross-section $A$ $(R \propto 1/A)$. When the area is doubled, resistance becomes half, and consequently, the current $I$ becomes double.
$(b)$ It means that $1\, J$ of work is being done to move a charge of $1\, C$ from point $A$ to point $B$.

(A) The circuit diagram is as shown in the provided image.
Given:
Total voltage $V = 5 \times 2 \text{ V} = 10 \text{ V}$.
Resistors in series: $R_{1} = 5 \, \Omega$, $R_{2} = 10 \, \Omega$, $R_{3} = 15 \, \Omega$.
The equivalent resistance $R$ of the series combination is:
$R = R_{1} + R_{2} + R_{3} = 5 + 10 + 15 = 30 \, \Omega$.
$(i)$ The electric current $I$ passing through the circuit is:
$I = V / R = 10 \text{ V} / 30 \, \Omega = 1/3 \text{ A} \approx 0.33 \text{ A}$.
$(ii)$ The potential difference $V_{1}$ across the $5 \, \Omega$ resistor is:
$V_{1} = I \times R_{1} = (1/3) \text{ A} \times 5 \, \Omega = 5/3 \text{ V} \approx 1.67 \text{ V}$.

(N/A) An electric circuit is a continuous and closed path in which an electric current can flow.
$(b)$ We know that $I = Q / t$ and $Q = n e$,where $n$ is the number of electrons and $e$ is the charge on an electron. Therefore,we have:
$I = n e / t$
$n = (I \times t) / e = (1 \ \text{A} \times 1 \ \text{s}) / (1.6 \times 10^{-19} \ \text{C})$
$n = 6.25 \times 10^{18}$ electrons.
$(c)$ The circuit diagram used to study Ohm's law is shown below.
Electric current is measured by an Ammeter $(A)$.
Potential difference is measured by a Voltmeter $(V)$.

(N/A) $Ohm's$ law: The potential difference across the ends of a resistor is directly proportional to the current flowing through it, provided its temperature remains constant.
Mathematical representation: $V = IR$ or $R = V / I$, where $V$ is the potential difference, $I$ is the current, and $R$ is the resistance.
Definition of $1\, ohm$: If the potential difference across the two ends of a conductor is $1\, V$ and the current flowing through it is $1\, A$, then the resistance $R$ of the conductor is $1\, ohm$ $(1\, \Omega = 1\, V / 1\, A)$.
Disadvantages of series connection:
$(i)$ Different electrical appliances require different values of current to operate properly, which is not possible in a series circuit because the current remains the same throughout.
$(ii)$ If one appliance fails or is switched off, the entire circuit is broken, and none of the other appliances will work.

(N/A) When three resistors $R_1, R_2,$ and $R_3$ are connected in series,the same current $I$ flows through each resistor. The potential difference across each resistor is $V_1 = IR_1, V_2 = IR_2,$ and $V_3 = IR_3$. The total potential difference $V$ is $V = V_1 + V_2 + V_3 = I(R_1 + R_2 + R_3)$. If $R_s$ is the equivalent resistance,then $V = IR_s$. Thus,$R_s = R_1 + R_2 + R_3$.
$(b)$ The given circuit shows three resistors connected in parallel with values $R_1 = 6 \Omega, R_2 = 10 \Omega,$ and $R_3 = 15 \Omega$.
For parallel connection,the equivalent resistance $R_p$ is given by:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
$\frac{1}{R_p} = \frac{1}{6} + \frac{1}{10} + \frac{1}{15}$
$\frac{1}{R_p} = \frac{5 + 3 + 2}{30} = \frac{10}{30} = \frac{1}{3}$
Therefore,$R_p = 3 \Omega$.

(A) Given:
Potential difference across $5 \, \Omega$ resistor $(V_1)$ = $10 \, V$
Potential difference across $R \, \Omega$ resistor $(V_2)$ = $6 \, V$
Resistance $R_1 = 5 \, \Omega$
$(i)$ By Ohm's law,the current $(I)$ through the $5 \, \Omega$ resistor is:
$I = \frac{V_1}{R_1} = \frac{10 \, V}{5 \, \Omega} = 2 \, A$
(ii) Since the two resistors are connected in series,the current remains the same throughout the circuit. Therefore,the current through resistor $R$ is also $2 \, A$.
(iii) Using Ohm's law for resistor $R$:
$R = \frac{V_2}{I} = \frac{6 \, V}{2 \, A} = 3 \, \Omega$
(iv) Since the resistors are in series,the total voltage $V$ is the sum of the potential differences across each resistor:
$V = V_1 + V_2 = 10 \, V + 6 \, V = 16 \, V$

(A) Given: Voltage $V = 5\,V$,Current $I = 500\,mA = 0.5\,A$,Time $t = 4\,\text{hours} = 4 \times 3600\,s = 14400\,s$.
$(i)$ Power $P = V \times I = 5\,V \times 0.5\,A = 2.5\,W$.
$(ii)$ Resistance $R = V / I = 5\,V / 0.5\,A = 10\,\Omega$.
$(iii)$ Energy consumed $E = P \times t = 2.5\,W \times 14400\,s = 36000\,J$.

(B) When electric lamps are connected in parallel,the total power consumed by the circuit is the sum of the power of individual lamps.
Total power,$P = P_{1} + P_{2} = 100\, W + 25\, W = 125\, W$.
The relationship between power $(P)$,voltage $(V)$,and current $(I)$ is given by the formula $P = V \times I$.
To find the total current $(I)$,we rearrange the formula: $I = \frac{P}{V}$.
Substituting the given values: $I = \frac{125\, W}{200\, V} = 0.625\, A$.
Therefore,the total current flowing through the circuit is $0.625\, A$.

(N/A) $(i)$ Since $R_{1}$ and $R_{2}$ are in series,their resultant resistance,
$R_{S} = R_{1} + R_{2} = 3 + 2 = 5 \Omega$
Further,$R_{S}$ and $R_{3}$ are in parallel,their resultant resistance is given by
$R_{P} = \frac{R_{S} \times R_{3}}{R_{S} + R_{3}} = \frac{5 \times 5}{5 + 5} = 2.5 \Omega$
(ii) Also,the current $I$ shown by the ammeter is given by Ohm's law:
$I = \frac{V}{R_{P}} = \frac{2.5 \text{ V}}{2.5 \Omega} = 1 \text{ A}$

(N/A) In arm $CE$, the resistors of value $2 \, \Omega$ and $4 \, \Omega$ are connected in series. Therefore, the total resistance in arm $CE = 2 \, \Omega + 4 \, \Omega = 6 \, \Omega$.
$(b)$ The circuit consists of two parallel branches, arm $CE$ and arm $AB$, each having a resistance of $6 \, \Omega$. The total resistance of the circuit $R_P$ is given by:
$R_P = \frac{6 \, \Omega \times 6 \, \Omega}{6 \, \Omega + 6 \, \Omega} = 3 \, \Omega$.
Therefore, the total current $I$ drawn from the battery is:
$I = \frac{V}{R_P} = \frac{3 \, V}{3 \, \Omega} = 1 \, A$.
$(c)$ Since both arms $CE$ and $AB$ have the same resistance $(6 \, \Omega)$, the total current is split equally between them. Hence, the current in each arm ($AB$ and $CE$) is $0.5 \, A$.

(N/A) The two $8\, \Omega$ resistors are connected in parallel. Their effective resistance $R_P$ is given by:
$R_P = \frac{R_1 R_2}{R_1 + R_2} = \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4\, \Omega$
$(b)$ The total resistance of the circuit $R_{eq} = R_{series} + R_P = 4\, \Omega + 4\, \Omega = 8\, \Omega$. The total current $I$ in the circuit is:
$I = \frac{V}{R_{eq}} = \frac{8\, V}{8\, \Omega} = 1\, A$. Since the $4\, \Omega$ resistor is in series with the rest of the circuit,the same current flows through it.
$(c)$ The potential difference $V$ across the $4\, \Omega$ resistor is:
$V = I \times R = 1\, A \times 4\, \Omega = 4\, V$
$(d)$ The power $P$ dissipated in the $4\, \Omega$ resistor is:
$P = I^2 R = (1\, A)^2 \times 4\, \Omega = 4\, W$
$(e)$ There is no difference in the ammeter readings. In a series circuit,the current remains the same at all points,so both ammeters $A_1$ and $A_2$ will show the same reading of $1\, A$.

(N/A) Given: Voltage $V = 200 \, V$,Resistance of coil $A$ $(R_A)$ = $30 \, \Omega$,Resistance of coil $B$ $(R_B)$ = $30 \, \Omega$.
$(i)$ When coil $A$ or $B$ is used separately:
$I = V / R = 200 / 30 = 6.67 \, A$.
(ii) When the coils are connected in series:
Total resistance $R_S = R_A + R_B = 30 + 30 = 60 \, \Omega$.
$I_S = V / R_S = 200 / 60 = 3.33 \, A$.
(iii) When the coils are connected in parallel:
Total resistance $1 / R_P = 1 / R_A + 1 / R_B = 1 / 30 + 1 / 30 = 2 / 30 = 1 / 15$.
$R_P = 15 \, \Omega$.
$I_P = V / R_P = 200 / 15 = 13.33 \, A$.

$(A)$ Sample $A$ is a good conductor, and sample $B$ is an insulator.
Metals and alloys typically have low resistivity in the range of $10^{-8} \ \Omega \cdot m$ to $10^{-6} \ \Omega \cdot m$, making them good conductors of electricity. In contrast, insulators possess very high resistivity, typically in the order of $10^{12} \ \Omega \cdot m$ to $10^{17} \ \Omega \cdot m$.
$(b)$ When a wire of resistance $R = 4 \ \Omega$ is folded on itself, its length $l$ becomes $l' = l/2$ and its cross-sectional area $A$ becomes $A' = 2A$.
The resistance formula is $R = \rho (l/A)$.
The new resistance $R'$ is given by:
$R' = \rho (l'/A') = \rho ((l/2) / (2A)) = (1/4) \times \rho (l/A) = R/4$.
Substituting $R = 4 \ \Omega$:
$R' = 4 \ \Omega / 4 = 1 \ \Omega$.
Therefore, the new resistance is $1 \ \Omega$.

(N/A) $(i)$ The net resistance of a series combination of resistors is greater than the net resistance of a parallel combination of resistors. Therefore,the resistance of circuit $I$ is greater than the resistance of circuit $II$.
$(ii)$ Since current is inversely proportional to resistance for the same potential difference $(I = V/R)$,the current in circuit $II$ is greater than the current in circuit $I$.
$(iii)$ In a parallel circuit,the potential difference across each resistor is equal. Therefore,the potential difference across each resistor in circuit $II$ is equal.

(N/A) Given: Rated voltage $V = 220 \, V$,Rated power $P = 60 \, W$.
$(i)$ Using the formula $P = \frac{V^2}{R}$,the resistance $R$ of the bulb is:
$R = \frac{V^2}{P} = \frac{(220)^2}{60} = \frac{48400}{60} \approx 806.67 \, \Omega$.
$(ii)$ When two identical bulbs are connected in series,the total resistance $R_T = R + R = 2R = 2 \times 806.67 = 1613.34 \, \Omega$.
The current $I$ flowing through the circuit is $I = \frac{V}{R_T} = \frac{220}{1613.34} \approx 0.136 \, A$.
The rate of energy conversion (power consumed) in each bulb is $P' = I^2 R = (0.136)^2 \times 806.67 \approx 15 \, W$.
$(iii)$ Total power consumed in series connection is $P_{total} = I^2 R_T = (0.136)^2 \times 1613.34 \approx 30 \, W$.
$(iv)$ If the bulbs are connected in parallel,the total power is $P_{total} = P_1 + P_2 = 60 + 60 = 120 \, W$.

(C) In electrical circuit diagrams,different components are represented by standard symbols.
- The symbol shown in option $A$ represents an open switch.
- The symbol shown in option $B$ represents a cell.
- The symbol shown in option $C$ represents a resistor,which is a component used to provide resistance in a circuit.
- The symbol shown in option $D$ represents a galvanometer,used to detect small electric currents.
Therefore,the correct symbol for a resistor is given in option $C$.

(A) $1$. Analyze the observations: The bulb glows $(ON)$ only if both materials connected at $X$ and $Y$ are conductors. If the bulb is $OFF$,it means at least one of the materials is an insulator.
$2$. From observation $2$,$I$ and $IV$ result in $ON$,so both $I$ and $IV$ are conductors.
$3$. From observation $4$,$III$ and $V$ result in $ON$,so both $III$ and $V$ are conductors.
$4$. From observation $1$,$I$ and $II$ result in $OFF$. Since $I$ is a conductor,$II$ must be an insulator.
$5$. For the third row,the bulb is $OFF$. This means at least one of the materials must be an insulator. Since $II$ is the only identified insulator,the pair must contain $II$.
$6$. Looking at the options,option $A$ ($II$ and $III$) contains $II$,which is an insulator. Therefore,the bulb will remain $OFF$.