Current Electricity

$\because V_{A B}=\left(I-I_g\right) R=I_g \cdot G$

where, G = voltmeter (resistance).

$\begin{array}{ll} \text { Given, } & V_{A B}=120 \mathrm{~V}, I=2 \mathrm{~A}, R=75 \Omega \\\\ \Rightarrow & 120=\left(2-I_g\right) 75 \Rightarrow I_g=0.4 \mathrm{~A} \\\\ \text { Now, } & V_{A B}=I_g G \Rightarrow G=\frac{120}{0.4}=300 \Omega \end{array}$

During charging, $E=V+i r$ (due to reversed current). In case of charging emf of a cell is less than its terminal voltage while in case of discharging emf is greater than terminal voltage.

Switching results in high decay/growth rate of current which results in a high. Current when bulb is turned off (due to back emf). So, a bulb is must likely to get fused when it is just turned off.

$\text { Given, } J_B=\frac{J_A}{2}$

$\begin{aligned} &\begin{aligned} \text { i.e., } 2 \times I\left(\pi r_A^2\right) & =I\left(\pi r_B^2\right) \\ \Rightarrow \quad r_B & =\left(r_A\right) \sqrt{2}=0.5 \times 1.414 \\ & =0.7 \mathrm{~cm} \end{aligned} \end{aligned}$

Figure is equivalent to the one shown below.

It is a Wheatstone’s bridge in which

$\begin{aligned} & P=\frac{6 \times 3}{6+3}=2 \Omega \\\\ & Q=\frac{8}{2}=4 \Omega \\\\ & R=\frac{4}{2}=2 \Omega \text { and } S=\frac{20 \times 5}{20+5}=4 \Omega \end{aligned}$

We find that $\frac{P}{Q}=\frac{2}{4}=\frac{R}{S}$

i.e. the bridge is balanced. Effective resistance between $p$ and $q$

$\begin{aligned} R_{A B} & =\frac{(P+Q)(R+S)}{P+Q+R+S}=\frac{(2+4)(2+4)}{2+4+2+4} \\\\ & =\frac{6 \times 6}{12}=3 \Omega \end{aligned}$

It is clear that the two cells oppose each other hence the effective emf in closed circuit is $18-12=6 \mathrm{~V}$ and net resistance is $1+2=3 \Omega$ (because in the closed circuit the internal resistance of two cells are in series)

The current in circuit will be in direction of arrow shown in figure

$\begin{aligned} I & =\frac{\text { effective emf }}{\text { total resistance }} \\\\ & =6 / 3=2 \mathrm{~A} \end{aligned}$

The potential difference across V will be same as the terminal voltage of either cell

Since, current is drawn from the cell of $18 \mathrm{~V}$ hence

$\begin{aligned} V_1 & =E_1-I r_1=18-(2 \times 2) \\\\ & =18-4=14 \mathrm{~V} \end{aligned}$

When a voltmeter is connected across the two terminals of a cell, it draws a small current from the cell, so it measures terminal potential difference between the two terminals of the cell, which is always less than the emf of the cell on the other hand, when a potentiometer is used for the measurement of emf of cell, it does not draw any current from the cell. Hence, it accurately measures the emf of cell. Thus, a potentiometer is preferred over a voltmeter.