Question

A battery of emf $12\text{V}$ having $10\Omega$ internal resistance is connected with $100\Omega$ resistor. If the resistance of a resistor increases by $1\%$, while total power loss remains constant, what should be the percentage change in emf ?

• A. $0.5\%$
• B. $0.45\%$
• C. $0.55\%$
• D. $0.62\%$

Answer 1

The correct option is B $0.45\%$

Given:
$E=12\text{V};R=100\Omega ;r=10\Omega$

Current flowing through the circuit is,

$I=\frac{E}{R+r}$

So, the total power dissipated from the circuit will be

$P_t=I^{2}R+I^{2}r=I^2(R+r)$

$P_t=\frac{E^2}{(R+r{)}^2}(R+r)$

$\therefore P_t=\frac{E^2}{(R+r)}$

Taking $\log$ of both sides,

$\log P_t=2\log E-\log (R+r)$

On differentiating, we get

$\frac{d{P}_t}{P_t}=\frac{2dE}{E}-\frac{d(R+r)}{(R+r)}$

Since, $P_t$ and $r$ are constants, so $d{P}_t=0$, and $dr=0$

$\therefore$ $\frac{2dE}{E}=\frac{dR}{(R+r)}.........\left(1\right)$

Given that, $dR=1\%\text{of}R=0.01\times 100=1\Omega$

$\therefore \frac{dR}{R+r}=\frac{1}{100+10}$

So, equation $\left(1\right)$ becomes

$2\times \frac{dE}{E}=\frac{1}{110}\Rightarrow \frac{dE}{E}=\frac{1}{220}$

$\therefore \frac{dE}{E}\times 100=\frac{1}{220}\times 100=0.45$

Thus, the percentage change in emf is $0.45\%$.