Question

For the circuit shown in the figure, the potential difference across each of the four batteries $B_1,B_2,B_3$ and $B_4$ respectively will be:

• A. $4\text{V},0\text{V},8\text{V},8\text{V}$
• B. $4\text{V},0\text{V},9\text{V},6\text{V}$
• C. $4\text{V},1\text{V},8\text{V},7\text{V}$
• D. $4\text{V},0\text{V},7\text{V},1\text{V}$

Answer 1

The correct option is D $4\text{V},0\text{V},7\text{V},1\text{V}$

From the circuit shown in the figure, we can observe that battery $B_1$ is connected in open branch, thus no current will flow through them.

So, potential difference across $B_1=\text{EMF of}{B}_1$

$\therefore V_1=4\text{V}$

Battery $B_2$ is short-circuited, hence the potential difference across will be zero.

$\therefore V_2=0$

Here, only battery $B_3$ and $B_4$ forms a closed circuit.


From above diagram,

$E_{net}=10+5=15\text{V}$

Current in circuit,

$i=\frac{E_{net}}{R_{eq}}=\frac{15}{2+2+1}=3\text{A}$

Thus, for potential difference across $B_3$,

$V_a-10+i\left(1\right)=V_d$ (From KVL)

Or, $V_a-V_d=10-i=10-3$

Or, $V_3=7\text{Volt}$

Similarly, for battery $B_4$,

$V_b-i\left(2\right)+5=V_c$

$\Rightarrow V_b-V_c=5-2i=5-6=-1$

Or, $V_c-V_b=1\text{V}$

$\therefore V_4=1\text{Volt}$

Hence, option $\left(d\right)$ is correct.