Question

The figure shows an infinite circuit formed by the repetition of resistance 4 $\Omega$ and 3 $\Omega$ and a cell of emf 18 V. Then

• A. Current through the cell is 3 A
• B. Current through the branch AC is 1 A

Answer 1

Answer: (A)

A. Current through the cell is 3 A

Answer 2

The circuit shown is an infinite ladder network. To find the equivalent resistance of such a network, we assume that the equivalent resistance of the entire infinite network is $R_{eq}$. Since the network is infinite, if we add one more identical repeating unit, the equivalent resistance remains the same.

1. Identify the repeating unit and set up the equation for equivalent resistance:
   The repeating unit consists of a 4 $\Omega$ resistor in series with a parallel combination of a 3 $\Omega$ resistor and the rest of the infinite network. Let the equivalent resistance of the entire infinite circuit be $R_{eq}$.
   Therefore, we can write the equation for $R_{eq}$ as:
   $R_{eq} = 4 \, \Omega + (3 \, \Omega \parallel R_{eq})$
   $R_{eq} = 4 + \frac{3 \times R_{eq}}{3 + R_{eq}}$

2. Solve for $R_{eq}$:
   Multiply both sides by $(3 + R_{eq})$:
   $R_{eq}(3 + R_{eq}) = 4(3 + R_{eq}) + 3 R_{eq}$
   $3R_{eq} + R_{eq}^2 = 12 + 4R_{eq} + 3R_{eq}$
   $3R_{eq} + R_{eq}^2 = 12 + 7R_{eq}$
   Rearrange into a quadratic equation:
   $R_{eq}^2 - 4R_{eq} - 12 = 0$
   Factor the quadratic equation:
   $(R_{eq} - 6)(R_{eq} + 2) = 0$
   This gives two possible values for $R_{eq}$: $R_{eq} = 6 \, \Omega$ or $R_{eq} = -2 \, \Omega$.
   Since resistance cannot be negative, we take the positive value:
   $R_{eq} = 6 \, \Omega$

3. Calculate the current through the cell (Option A):
   The total equivalent resistance of the circuit connected to the 18 V cell is $R_{eq} = 6 \, \Omega$.
   Using Ohm's Law, the current through the cell ($I_{cell}$) is:
   $I_{cell} = \frac{\text{EMF}}{R_{eq}} = \frac{18 \, V}{6 \, \Omega} = 3 \, A$
   So, statement A is correct.

4. Calculate the current through the branch AC (Option B):
   The current $I_{cell} = 3 \, A$ flows through the first 4 $\Omega$ resistor.
   The voltage drop across this first 4 $\Omega$ resistor is:
   $V_{4\Omega} = I_{cell} \times 4 \, \Omega = 3 \, A \times 4 \, \Omega = 12 \, V$
   The voltage across the parallel combination (which includes the branch AC, the 3 $\Omega$ resistor) is the total EMF minus the voltage drop across the first 4 $\Omega$ resistor. This voltage is $V_{AC}$.
   $V_{AC} = 18 \, V - 12 \, V = 6 \, V$
   Now, the current through the branch AC (the 3 $\Omega$ resistor) is:
   $I_{AC} = \frac{V_{AC}}{3 \, \Omega} = \frac{6 \, V}{3 \, \Omega} = 2 \, A$
   So, statement B, which states the current through branch AC is 1 A, is incorrect.