Question

Two resistors $400\Omega$ and $800\Omega$ are connected in series across a $6V$battery. The potential difference measured by a voltmeter of $10k\Omega$ across $400\Omega$ resistors is close to:

• A. $2.05V$
• B. $2V$
• C. $1.95V$
• D. $1.8V$

Answer 1

The correct option is C

$1.95V$

Step 1: Given

The resistance of the two resistors is $400\Omega$ and $800\Omega$.

The potential difference is $6V$.

The value of the resistance of the voltmeter added across $400\Omega$ resistor is $10k\Omega$.

Step 2: Solution

Let the total resistance of $10k\Omega$ and $400\Omega$ be $R$, the equivalent resistance be $R_{eq}$ and the current flowing through the circuit be $i$.

Let the potential difference across the $400\Omega$ resistors be $V_{400}$.

The required total resistance between parallel resistors of $10k\Omega$ and $400\Omega$ is

$$\begin{gathered} \frac{1}{R}=\frac{1}{10000}+\frac{1}{400}\left[Inparallelconnection\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\right] \\ \Rightarrow R=\frac{400\times 10000}{10400} \\ \Rightarrow R=\frac{40000}{104}=384.6153\Omega \end{gathered}$$

The equivalent resistance of the circuit

$$\begin{gathered} R_{eq}=R+800\left[Inseriesresis\tan ceR=R_1+R_2\right] \\ \Rightarrow R_{eq}=384.6153+800 \\ \Rightarrow R_{eq}=1184.6153\Omega \end{gathered}$$

The value of the current $i$ is

$$\begin{gathered} i=\frac{V}{R_{eq}}[Byohm\text{'}slaw] \\ \Rightarrow i=\frac{6}{1184.6153} \\ \Rightarrow i=0.00506A \end{gathered}$$

The potential difference across $400\Omega$ resistors is

$$\begin{gathered} V_{400}=6-\left(800\times 0.00506\right) \\ \Rightarrow V_{400}=6-4.048 \\ \Rightarrow V_{400}=1.952V \end{gathered}$$

Hence the correct option is $C$.