Question 121: Current Electricity

Question

A meter bridge with two resistances $R_1$ and $R_2$ as shown in figure was balanced (null point) at 40 cm from the point $P$. The null point changed to 50 cm from the point $P$, when $16 \Omega$ resistance is connected in parallel to $R_2$. The values of resistances $R_1$ and $R_2$ are $\_\_\_\_$ .

Accepted Answer

A meter bridge operates on the principle of a Wheatstone bridge.
Let the total length of the wire be $\mathrm{L}=100 \mathrm{~cm}$.
If the null point is found at a distance l from the left end, the length of the right segment is $(100-\mathrm{l})$.
The balancing condition is given by:
$$ \frac{\mathrm{R}_{	ext {left gap }}}{\mathrm{R}_{	ext {right gap }}}=\frac{\mathrm{l}}{100-\mathrm{l}} $$
The resistance in left gap is $R_{	ext {left gap }}=R_1$ and the resistance in the right gap is $R_{	ext {right gap }}=R_2$
Case 1
We are given that initially; the bridge is balanced at a distance of $\mathrm{l}_1=40 \mathrm{~cm}$ from point P .
The remaining length is $100-40=60 \mathrm{~cm}$.
Applying the balancing condition :
$$ \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{40}{60} $$
$$\Rightarrow $$ $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{2}{3}$
$$\Rightarrow $$ $$ \mathrm{R}_1=\frac{2}{3} \mathrm{R}_2 \ldots (i)$$
Case 2
When a $16 \Omega$ resistance is connected in parallel to $R_2$.
So, the new resistance in the right gap be $\mathrm{R}_2^{\prime}$ :
$$ \frac{1}{\mathrm{R}_2^{\prime}}=\frac{1}{\mathrm{R}_2}+\frac{1}{16} \Rightarrow \mathrm{R}_2^{\prime}=\frac{16 \mathrm{R}_2}{16+\mathrm{R}_2} $$
With this new resistance, the null point shifts to $\mathrm{l}_2=50 \mathrm{~cm}$ from point P .
The remaining length is now $100-50=50 \mathrm{~cm}$.
Applying the balancing condition again,
$$ \frac{\mathrm{R}_1}{\mathrm{R}_2^{\prime}}=\frac{50}{50}=1 $$
$$ \mathrm{R}_1=\mathrm{R}_2^{\prime} \ldots (ii)$$
Substituting the expression for $\mathrm{R}_2^{\prime}$ into equation (ii):
$$ R_1=\frac{16 R_2}{R_2+16} \ldots 	ext { (iii) } $$
Now, substituting the expression for $R_1$ from equation into equation (iii) :
$$ \frac{2}{3} R_2=\frac{16 R_2}{R_2+16} $$
$$\Rightarrow $$ $\frac{1}{3}=\frac{8}{R_2+16} \Rightarrow R_2+16=24$
$$\Rightarrow $$ $$ \mathrm{R}_2=8 \Omega $$
And, $\mathrm{R}_1=\frac{2}{3} 	imes 8 \Omega=\frac{16}{3} \Omega$
Therefore, the values are $\mathrm{R}_2=8 \Omega$ and $\mathrm{R}_1=\frac{16}{3} \Omega$.
Hence, the correct option is (A).

Answer

$R_2=8 \Omega, R_1=\frac{16}{3} \Omega$

Answer

$R_2=12 \Omega, R_1=\frac{12}{3} \Omega$

Answer

$R_2=4 \Omega, R_1=\frac{4}{3} \Omega$

Answer

$R_2=16 \Omega, R_1=\frac{16}{3} \Omega$

Current Electricity

A meter bridge with two resistances R_1 and R_2 as shown in figure was balanced (null point) at 40 cm from the point P. The null point changed to 50 cm from the point P, when 16 \Omega resistance is connected in parallel to R_2. The values of resistances R_1 and R_2 are \_\_\_\_ .

Choose the correct answer: