Question

A battery having emf $E_1=100\text{V}$ and internal resistance $r=0.5\Omega$ is connected to a storage battery of emf $E_2=90\text{V}$ and external resistance $R$ as shown in figure. For what value of $R$ no current will pass through storage battery $E_2$?

• A. $7.5\Omega$
• B. $4.5\Omega$
• C. $2.5\Omega$
• D. $5.5\Omega$

Answer 1

The correct option is B $4.5\Omega$

Let that no current passes through battery $E_2$, therefore the current supplied by power supply $E_1$ will flow through resistance $R$ in a closed circuit.


For the loop $\text{ABCDA}$,

$i=\frac{E_1}{R+r}$

$\Rightarrow i=\frac{100}{R+0.5}.........\left(1\right)$

Now potential at points $\text{E}$ and $\text{F}$ can be related by KVL,

$V_E-iR=V_F$

$V_E-V_F=iR$

Also the potential difference across $\text{E}$ and $\text{F}$ will same as emf of battery $E_2$.

$\therefore E_2=iR$

$\Rightarrow i=\frac{E_2}{R}=\frac{90}{R}.....\left(2\right)$

From Eq.(1) and (2),

$\frac{90}{R}=\frac{100}{R+0.5}$

$\Rightarrow 90R+45=100R$

$\therefore R=4.5\Omega$

Hence, option $\left(b\right)$ is correct.
$$\begin{array}{c}\text{Why this question ?}\\ \text{It intends to test your understanding of}\\ \text{flow of current in a closed loop \& Ohm's}\\ \text{law across a component element/resistance.}\end{array}$$