Question

Find potential difference across the resistance $300\Omega$ given in circuit $\text{A}$ and $\text{B}$.

• A. $60\text{V}$ and $50\text{V}$
• B. $70\text{V}$ and $70\text{V}$
• C. $30\text{V}$ and $30\text{V}$
• D. $40\text{V}$ and $30\text{V}$

Answer 1

The correct option is A $60\text{V}$ and $50\text{V}$

In $\left(A\right)$:

Both the resistors are in series thus, potential difference across them directly proportional to their resistances.

$V_{\left(300\Omega \right)}=IR=\frac{100}{200+300}\times 300=60\text{V}$

In $\left(B\right)$:

$300\Omega$ and voltmeter are in parallel, their effctive resistance is in series with $200\Omega$

Total resistance $R_T$ is,

$R_T=200+\frac{300\times 600}{300+600}=400\Omega$

Total current through the battery is,

$I=\frac{100}{400}=0.25\text{A}$

As we know, across the parallel branches current gets divided in the inverse ratio of resistance of the branch.

$I_1:I_2=\frac{1}{300}:\frac{1}{600}=2:1$

$I_1=\frac{2}{3}\times 0.25=0.16\text{A}$

Hence potential difference across the $300\Omega$ is,

$V=0.16\times 300=50\text{V}$

Alternate Solution:

$\because 300\Omega$ and $600\Omega$ are in parallel, thus potential difference across them will be the same.

$V_{\left(300\Omega \right)}=IR=\frac{100}{200+\frac{300\times 600}{300+600}}\times \left[\frac{300\times 600}{300+600}\right]=50\text{V}$

Hence, option (a) is the correct answer.

Why this question ? We see that by the connected voltmeter the voltage which was to be measured has changed. Such voltmeters are not good. If its resistance had been very large than $300\Omega$ then it would have not affected the voltage by much amount.