The problem asks for the reading of an ideal voltmeter connected in the circuit as shown. The circuit consists of several cells, each with an EMF of 5V and an internal resistance of 0.2Ω.
Concept Used:
1. Ideal Voltmeter: An ideal voltmeter has infinite internal resistance. When connected to a circuit, it draws no current. Therefore, the branch containing the ideal voltmeter can be treated as an open circuit.
2. Kirchhoff's Voltage Law (KVL): The algebraic sum of the changes in electric potential around any closed circuit loop is zero. We will use KVL to find the current flowing in the circuit.
3. Terminal Voltage: The potential difference across the terminals of a cell is given by \(V = E - Ir\) when the cell is discharging (current flows out of the positive terminal) and \(V = E + Ir\) when it is charging (current flows into the positive terminal).
Step-by-Step Solution:
Step 1: Analyze the circuit based on the properties of the ideal voltmeter.
Since the voltmeter is ideal, it has infinite resistance. This means no current flows through the central branch where the voltmeter is located. The circuit effectively becomes a single closed loop consisting of the outer perimeter.
Step 2: Calculate the net Electromotive Force (EMF) in the outer loop.
Let's traverse the loop in the clockwise direction. We sum the EMFs of all the cells in the loop, considering their polarity.
Top branch: Three cells are in series, and their positive terminals are on the right. Total EMF = \(3 \times 5V = 15V\). This EMF drives current clockwise.
Left branch: One cell with its positive terminal at the top. This EMF also drives current clockwise. EMF = \(5V\).
Bottom branch: Three cells are in series, with their positive terminals on the right. This EMF also drives current clockwise. Total EMF = \(3 \times 5V = 15V\).
Since all EMFs support a clockwise current, the net EMF of the loop is the sum:
The voltmeter measures the potential difference between the top-right corner (let's call it point A) and the bottom-right corner (point B), i.e., \(V_{AB} = V_A - V_B\). We can calculate this potential difference by traversing any path from B to A and summing the potential changes.
Let's choose the path along the top branch of the circuit. To do this, let's first establish the potential at the nodes. Let's assume the potential at point B is zero, \(V_B = 0\). We will find the potential at A by going from B around the loop via the bottom, left, and top branches.
Potential at the bottom-left corner (D): Moving from B to D (right to left), we go against the current (\(I=25A\)) and against the EMF of the bottom branch (\(E_{bottom}=15V\)). The potential change across a resistor against current is a rise (\(+IR\)). The potential change against an EMF (from + to -) is a drop (\(-E\)). \[ V_D = V_B + I \cdot r_{bottom} - E_{bottom} = 0 + (25 \, \text{A})(0.6 \, \Omega) - 15 \, \text{V} = 15 \, \text{V} - 15 \, \text{V} = 0 \, \text{V} \]
Potential at the top-left corner (C): Moving from D to C (upwards), we go with the current and with the EMF. Potential drops across the resistor (\(-IR\)) and rises across the EMF (\(+E\)). \[ V_C = V_D - I \cdot r_{left} + E_{left} = 0 - (25 \, \text{A})(0.2 \, \Omega) + 5 \, \text{V} = -5 \, \text{V} + 5 \, \text{V} = 0 \, \text{V} \]
Potential at the top-right corner (A): Moving from C to A (left to right), we go with the current and with the EMF. \[ V_A = V_C - I \cdot r_{top} + E_{top} = 0 - (25 \, \text{A})(0.6 \, \Omega) + 15 \, \text{V} = -15 \, \text{V} + 15 \, \text{V} = 0 \, \text{V} \]
The potential at point A is 0 V, and the potential at point B is 0 V.
Final Computation & Result:
The reading of the voltmeter is the potential difference between points A and B.
