Question

If the ammeter reading in the given circuit is zero, find the value of the resistance R. 1

Answer 1

150 \Omega

Answer 2

The problem describes a circuit similar to a Wheatstone bridge but with voltage sources in some arms. The key information is that the ammeter reading is zero. An ammeter placed in a branch measures the current flowing through that branch. If the reading is zero, it implies that no current flows through the ammeter.

1. Identify the nodes and set potentials:
   Let the nodes be A, B, C, and D as labeled in the diagram.
   The ammeter is connected between nodes A and D.
   Given that the ammeter reading is zero, it means there is no current flowing between A and D. This implies that the potential difference between A and D is zero: $V_A - V_D = 0$, or $V_A = V_D$.
   For simplicity, let's set the potential of node D to zero (ground): $V_D = 0 \, \text{V}$.
   Therefore, $V_A = 0 \, \text{V}$.

2. Determine potentials of other nodes using voltage sources:

   • Branch AD: This branch contains the ammeter. Since $V_A = V_D = 0 \, \text{V}$, no current flows through the ammeter, which is consistent with the given condition.
   • Branch AB: There is a 10V battery between A and B. The positive terminal is at B and the negative terminal is at A (indicated by the longer line for positive and shorter line for negative).
     So, $V_B - V_A = 10 \, \text{V}$.
     Since $V_A = 0 \, \text{V}$, we have $V_B = 10 \, \text{V}$.
   • Branch CD: There is a 4V battery between C and D. The positive terminal is at C and the negative terminal is at D.
     So, $V_C - V_D = 4 \, \text{V}$.
     Since $V_D = 0 \, \text{V}$, we have $V_C = 4 \, \text{V}$.

3. Calculate currents through resistors:
   Now we have the potentials of all nodes: $V_A = 0 \, \text{V}$, $V_B = 10 \, \text{V}$, $V_C = 4 \, \text{V}$, $V_D = 0 \, \text{V}$.

   • Current through the 100 $\Omega$ resistor (branch AC):
     The current $I_{CA}$ flows from C to A because $V_C > V_A$.
     $I_{CA} = \frac{V_C - V_A}{100 \, \Omega} = \frac{4 \, \text{V} - 0 \, \text{V}}{100 \, \Omega} = \frac{4}{100} \, \text{A} = 0.04 \, \text{A}$.
   • Current through the R resistor (branch BC):
     The current $I_{BC}$ flows from B to C because $V_B > V_C$.
     $I_{BC} = \frac{V_B - V_C}{R} = \frac{10 \, \text{V} - 4 \, \text{V}}{R} = \frac{6}{R} \, \text{A}$.

4. Apply Kirchhoff's Current Law (KCL) at a node:
   Since the ammeter reading is zero, the branch AD effectively acts as an open circuit for current flow from the main bridge. This means that the current flowing into node A from branch CA must be equal to the current flowing out of node A into branch AB (through the 10V battery).
   Let's apply KCL at node A:
   Current entering A = Current leaving A
   $I_{CA} = I_{AB}$ (current flowing from A to B through the 10V source).

Similarly, at node C:
Current entering C = Current leaving C
$I_{BC} + I_{DC} = I_{CA}$ (where $I_{DC}$ is current from D to C through the 4V source).

Consider the loop ABCDA. Since the ammeter current is zero, the current flowing through the 10V source and R must be the same as the current flowing through the 100 Ohm resistor and the 4V source.
More precisely, consider the junction B. The current $I_{AB}$ (from A to B) flows into B. The current $I_{BC}$ (from B to C) flows out of B. Since there are no other branches connected to B, by KCL at node B:
$I_{AB} = I_{BC}$.

Now, substitute the values we found:
From KCL at node A: $I_{CA} = I_{AB} = 0.04 \, \text{A}$.
From KCL at node B: $I_{AB} = I_{BC}$.
Therefore, $I_{BC} = 0.04 \, \text{A}$.

We also know that $I_{BC} = \frac{6}{R}$.
So, $\frac{6}{R} = 0.04$.
$R = \frac{6}{0.04} = \frac{6}{4/100} = \frac{6 \times 100}{4} = \frac{600}{4} = 150 \, \Omega$.

The value of the resistance R is $150 \, \Omega$.