(a) Find the equivalent resistance between points A and B in the circuit shown in figure.

(b) Find the value of current I in the circuit shown in figure.

OR

(i) The length of a potentiometer wire is 600 cm and it carries a current of 40 mA. For a cell of emf 2V and internal resistance 10 W, the balancing length is found to be 500 cm. If a voltmeter is connected across the cell, the balancing length is decreased by 10 cm. Find

(a) the resistance of the potentiometer wire,

(b) the resistance of voltmeter and

(c) the reading of the voltmeter.

 

(ii) A car battery has e.m.f. 12 volt and internal resistance 5 × 10^–2 ohm. If it draws 60 amp current, what will be the terminal voltage (in volt) of the battery?

(a) The network of resistances shown in question is not a balanced Wheatstone's bridge.

In such a case, to find the resistance between A and B, we connect a battery of voltage V across A and B and find the current 1 drawn from the battery in terms of V and find the equivalent resistance from the relation V = JR.

The currents in various branches are shown above.

Applying loops rule to loops AFBGHA, ACDFA and DEBFD we have

$\begin{array}{r}V-1\left(I-I_1\right)-2\left(I-I_1+I_2\right)=0\\ 2I_1+I_2-\left(I-I_1\right)=0\\ \left(I_1-I_2\right)+2\left(I-I_1+I_2\right)-I_2=0\end{array}$

Simplifying these equations we get

$\begin{array}{r}3I-3I_1+2I_2=V\\ I-3I_1-I_2=0\end{array}$

and

$2I-3I_1+4I_2=0$

Eliminating ${\mathrm{I}}_1$ from (v) and (vi), we have ${\mathrm{I}}_2=-\frac{\mathrm{I}}{5}$.

Using this value of ${\mathrm{I}}_1$ in (v) we get ${\mathrm{I}}_1=\frac{2}{5}\mathrm{I}$.

Using these values of ${\mathrm{I}}_1$ and $I_2$ in (iv) we get

$7\mathrm{I}=5\mathrm{V}\Rightarrow \frac{\mathrm{V}}{\mathrm{I}}=\frac{7}{5}=1.4\Rightarrow {\mathrm{R}}_{\mathrm{AB}}=1.4\mathrm{\Omega }$

(b) V_A - V_D = - 4I – 3I + 9 – 2I

= -9I + 9

But V_A = V_D because points A and D are earthed.

- 9I + 9 = 0 => I = l A

OR

(i) AB = 600cm, I = 40 mA = 0.04 A,

AC= 500 cm and AC' = 490 cm.

Let r be the resistance of AB.

(a) Resistance per cm of $\mathrm{AB}=\frac{\mathrm{r}}{600}\mathrm{\Omega }{\mathrm{cm}}^{-1}$. 

Therefore, the resistance of AC is ${\mathrm{R}}_{\mathrm{AC}}=\frac{\mathrm{r}}{600}\times 500=\frac{5\mathrm{r}}{6}\mathrm{\Omega }$

$\therefore$ Potential difference across AC due to driver cell is ${\mathrm{V}}_{\mathrm{AC}}={\mathrm{IR}}_{\mathrm{AC}}=0.04\times \frac{5\mathrm{r}}{6}\text{ volt }$

When the voltmeter is not connected, the potential difference across AC due to cell of emf E is 2V (because no current is drawn from it at the balance point). Hence 

$\begin{array}{c}{\mathrm{V}}_{\mathrm{AC}}=2\\ \Rightarrow \frac{0.04\times 5\mathrm{r}}{6}=2\Rightarrow \mathrm{r}=60\mathrm{\Omega }\end{array}$

(b) Let R be the resistance of voltmeter. When it is connected across E, the current drawn from the cell of emf 2V is

${\mathrm{I}}^{\mathrm{\text{'}}}=\frac{2}{\mathrm{R}+10}$

Potential difference across the voltmeter is

${\mathrm{V}}_{\mathrm{R}}={\mathrm{I}}^{\mathrm{\text{'}}}\mathrm{R}=\frac{2\mathrm{R}}{\mathrm{R}+10}$

V_R must be equal to the potential difference across AC' due to the driver cell.

V_R = resistance of AC' x current I

$=\frac{60\times 490}{600}\times 0.04=1.96$

$\Rightarrow \mathrm{ }\frac{2R}{(R+10)}=1.96\Rightarrow R=490\mathrm{\Omega }$

(c) Voltmeter reading is

$V_R=\frac{2R}{(R+10)}=\frac{2\times 490}{(490+10)}=1.96\mathrm{V}$

(ii) E = V + Ir

V = 12 – 3 = 9 volt