Question 110: Current Electricity

Question

A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances ($R_1 = R_2 = R_3 = R_4$). When $R_3$ resistance is heated to some temperature, its resistance value has gone up by 10%. The potential difference ($V_a - V_b$) (after $R_3$ is heated) is _________ V.

Accepted Answer

Initially : $\mathrm{R}_1=\mathrm{R}_2=\mathrm{R}_3=\mathrm{R}_4=\mathrm{R}$.
After heating $\mathrm{R}_3$ :
1. $\mathrm{R}_1=\mathrm{R}$
2. $\mathrm{R}_2=\mathrm{R}$
3. $\mathrm{R}_3=\mathrm{R}+10 \%$ of $\mathrm{R}=1.1 \mathrm{R}$
4. $\mathrm{R}_4=\mathrm{R}$
Supply voltage $\mathrm{V}=40 \mathrm{~V}$.
The voltage across $R_1$ and $R_2$ is 40 V . So the current in this branch is $i_a=\frac{40 \mathrm{~V}}{R_1+R_2}$.
The voltage across $\mathrm{R}_1=\mathrm{V}_1=\mathrm{i}_{\mathrm{a}} \mathrm{R}_1=\left(\frac{40 \mathrm{~V}}{\mathrm{R}_1+\mathrm{R}_2}\right) \mathrm{R}_1=\left(\frac{40 \mathrm{~V}}{\mathrm{R}+\mathrm{R}}\right) \mathrm{R}=20 \mathrm{~V}$
The voltage across $\mathrm{R}_2=\mathrm{V}_2=\mathrm{i}_{\mathrm{a}} \mathrm{R}_2=\left(\frac{40 \mathrm{~V}}{\mathrm{R}_1+\mathrm{R}_2}\right) \mathrm{R}_2=\left(\frac{40 \mathrm{~V}}{\mathrm{R}+\mathrm{R}}\right) \mathrm{R}=20 \mathrm{~V}$
Assuming the potential of negative terminal of battery as 0 V then $\mathrm{v}_{\mathrm{a}}=\mathrm{V}_2=20 \mathrm{~V}$
Similarly for the branch containing $\mathrm{R}_3$ and $\mathrm{R}_4$,
The voltage across $R_3$ and $R_4$ is 40 V . So the current in this branch is $i_b=\frac{40 \mathrm{~V}}{R_3+R_4}$.
The voltage across $R_3=V_3=i_b R_3=\left(\frac{40 \mathrm{~V}}{R_3+R_4}\right) R_3=\left(\frac{40 \mathrm{~V}}{1.1 R+R}\right)(1.1 R)=20.95 \mathrm{~V}$
The voltage across $\mathrm{R}_4=\mathrm{V}_4=\mathrm{i}_{\mathrm{b}} \mathrm{R}_4=\left(\frac{40 \mathrm{~V}}{\mathrm{R}_3+\mathrm{R}_4}\right) \mathrm{R}_4=\left(\frac{40 \mathrm{~V}}{1.1 \mathrm{R}+\mathrm{R}}\right) \mathrm{R}=19.05 \mathrm{~V}$
Assuming the potential of negative terminal of battery as 0 V then $\mathrm{v}_{\mathrm{b}}=\mathrm{V}_4=19.05 \mathrm{~V}$
So, the potential difference between points a and b is,
$$ \left(\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}\right)=20 \mathrm{~V}-19.05 \mathrm{~V}=0.95 \mathrm{~V} $$
Therefore, the potential difference is 0.95 V . Hence, the correct option is (B).

Answer

1.05

Answer

0.95

Answer

0

Answer

2

Current Electricity

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