Solution:

Ans.

$$ R_{1}: R_{2}: R_{3}=1: 2: 3 $$

Let,

$$ \begin{aligned} & R_{1}=R \ & R_{2}=2 R \ & R_{3}=3 R \end{aligned} $$

Total resistance $\left(R_{T}\right)=R+2 R+3 R$

$$ \begin{array}{llrl} & & =48 \Omega \ \therefore & & R & =8 \Omega \ \text { Hence } & & R_{1} & =8 \Omega \ & & R_{2} & =16 \Omega \ \text { and } & & R_{3} & =24 \Omega \end{array} $$

According to circuit :

$$ \begin{aligned} & E=I\left(R_{T}+r\right) \ & 8=I\left(R_{T}+r\right) \ & 8=I(12+1) \ & I=\frac{8}{13} \mathrm{~A} \ & I=0.615 \mathrm{~A} \ & 8 \end{aligned} $$

Let current across $R_{1}$ is $I_{1}$

and current across $R_{2}$ is $I_{2}$

So current across

$$ I_{2}=I_{1} $$

$$ R_{3}=I_{3}=\left(I-I_{1}\right) $$

Apply Kirchhoff’s $2^{\text {nd }}$ Law

$$ \begin{aligned} I_{1} & =\frac{R_{3}}{\left(R_{1}+R_{2}+R_{3}\right)} \times \mathrm{I} \ & =\frac{24}{(8+16+24)} \times 0.615 \quad 1 / 2 \ I_{2} & =I_{1}=0.3075 \mathrm{~A} \ I_{3} & =I-I_{1} \ & =0.615-0.3075 \ & =0.3075 \mathrm{~A} \end{aligned} $$

[AT Q. 5. The network $P Q R S$, shown in the circuit diagram, has the batteries of $4 \mathrm{~V}$ and $5 \mathrm{~V}$ and negligible internal resistance. A milliammeter of $20 \Omega$ resistance is connected between $P$ and $R$. Calculate the reading in the milliammeter.

A [O.D. Comptt. I, II, III 2012]

Ans. Let us redraw circuit as shown.

Using Kirchhoff’s voltage law in closed loop $D A B C D$

$$ \begin{equation*} \left(I_{1}+I_{2}\right) R_{3}+I_{1} R_{1}-E_{1}=0 \tag{i} \end{equation*} $$

In a closed loop $A B F E A$

$$ \begin{equation*} \left(I_{1}+I_{2}\right) R_{3}+I_{2} R_{2}-E_{2}=0 \tag{ii} \end{equation*} $$

Substituting the values of $R_{1}, R_{2}, R_{3}, E_{1} & E_{2}$ in equations (i) & (ii)

We have $\quad 44 I_{1}+4 I_{2}=1$

$$ \text { and } \quad 5 I_{1}+20 I_{2}=1 $$

Solving above equations, we get

$$ I_{1}=0.0186 \mathrm{~A} & I_{2}=0.0454 \mathrm{~A} $$

Now, current in milliammeter

$$ \begin{aligned} I_{m} & =I_{1}+I_{2} \ & =0 \cdot 0186 \mathrm{~A}+0 \cdot 0454 \mathrm{~A} \ & =0 \cdot 064 \mathrm{~A}=64 \mathrm{~mA} \end{aligned} $$