Question

A battery of emf $12V$ and internal resistance $2\Omega$ is connected with two resistors A and B of resistance $4\Omega$ and $6\Omega$ respectively joined in series.
Find:
(i) Current in the circuit.
(ii) The terminal voltage of the cell.
(iii) The potential difference across $6\Omega$ Resistor.
(iv) Electrical energy spent per minute in $4\Omega$ Resistor.

Answer 1

$\text{Step 1: Make circuit diagram and apply KVL [Refer Fig. 1]}$

While applying KVL in anti-clockwise direction, increase in voltage is considered as positive and decrease in voltage as negative.

$12-4i-6i-2i=0$

$\Rightarrow I=1A$

$\text{Step 2: Find terminal voltage of the cell [Refer Fig. 2]}$

$V_{\text{terminal}}=E-ir$

$=12-2i=12-2\times 1=10V$

$\text{Step 3: Apply Ohm's Law for}6\Omega \text{Resistance}$ $\text{[Refer Fig. 3]}$

$V_B=i{R}_B=6i=6\times 1=6V$

$\text{Step 4: Energy spent in}4\Omega \text{Resistance}$ $\text{[Refer Fig. 4]}$

Energy Spent in 1 minute($t=60sec)$:

$E_A=i^2{R}_{A}t=(1A{)}^2\times 4\Omega \times 60s$

$=240J$