Two resistor each with resistance of $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\Omega }$ are connected in series to a 60 V power supply whose internal resistance is negligible. You connect the voltmeter across one of these resistors and this voltmeter has an internal resistance of $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\Omega }$. What is the reading on the voltmeter?

The potential of battery is $V=60\text{V}$.

The short circuit current of battery is $I_{SC}=12\text{A}$.

The resistance of each resistor in series is $R=4\times {10}^6\Omega$.

The internal resistance of the voltmeter is $r=1\times {10}^6\Omega$

The number of resistors in series isrole="math" localid="1668594014869" $n=2.$

The current in circuit without voltmeter is calculated on the basis of equivalent resistance of series resistors, then internal resistance of voltmeter is considered in parallel with one resistor to find the reading of voltmeter.

The current for the battery is given as:

$I=\frac{V}{nR}$

Substitute all the values in the above equation.

$$\begin{gathered} I=\frac{60\text{V}}{2\left(4\times {10}^6\Omega \right)} \\ I=7.5\times {10}^{-6}\text{A} \end{gathered}$$

The equivalent resistance for voltmeter reading is given as:

$R_e=\frac{rR}{r+R}$

Substitute all the values in the above equation.

$\begin{array}{rcl}{R}_e& =& \frac{\left(1\times {10}^6\Omega \right)\left(4\times {10}^6\Omega \right)}{\left(1\times {10}^6\Omega +4\times {10}^6\Omega \right)}\\ R_e& =& 0.8\times {10}^6\Omega \\ & & \end{array}$

The reading of the voltmeter is given as:

$E=I{R}_e$

Substitute all the values in the above equation.

$$\begin{gathered} E=\left(7.5\times {10}^{-6}\text{A}\right)\left(0.8\times {10}^6\Omega \right) \\ E=6\text{V} \end{gathered}$$

Therefore, the reading of the voltmeter is $6\text{V}$.