Two resistors are connected in series as shown in the diagram. $(i)$ What is the current through the $5 \, \Omega$ resistor? (ii) What is the current through $R$? (iii) What is the value of $R$? (iv) What is the value of $V$?
(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:
$V = V_1 + V_2 = 10 \, V + 6 \, V = 16 \, V$
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:
$V = V_1 + V_2 = 10 \, V + 6 \, V = 16 \, V$
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