Question

In the shown circuit the resistance $R$ can be varied. In which of the option, the variation of the current $i$ against $R$ is correctly plotted?

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Step 1: Kirchhoff's Current Law (KCL)

1. It states that the algebraic sum of the currents meeting at a junction point in a network is zero.
2. The law is a consequence of the conservation of charge.
3. In the steady-state, the rate at which a charge enters a junction must be equal to the rate at which it leaves the junction.
4. Mathematically the law may be stated as $\sum _k{I}_k=0$.
5. If the current entering a junction is taken as positive then the current leaving the junction should be taken as negative.

Step 2: Estimate Current vs Resistance graph

Upon applying Kirchhoff's law in this circuit we get,

$I_1+I_2+I_3=0$

Let $V$ be the total emf of the circuit.

$\therefore \frac{V-E}{r}+\frac{V-E}{r}+\frac{V-E}{R}=0$

$\Rightarrow \left(V-E\right)\left(\frac{2}{r}+\frac{1}{R}\right)=0$

From the given circuit we know that $\left(\frac{2}{r}+\frac{1}{R}\right)\ne 0$

Therefore, $\left(V-E\right)=0$

The current through $R$,

$i=\frac{V-E}{r}=\frac{0}{R}=0$

Therefore, the graph will be along $R$-axis.

Hence, the correct answer is option (C).