To determine the terminal potential difference of the cell in the given circuit, let's analyze the components and apply the necessary formulas.
From the circuit, the cell has an electromotive force (emf) \(E = 3\, V\) and an internal resistance \(r = 1\, \Omega\).
The resistors in the circuit are \(4\, \Omega\) each, and they are connected in parallel.
The equivalent resistance \(R\) of the parallel resistors can be calculated using: \(\frac{1}{R} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\)\(\Rightarrow R = 2\, \Omega\)
The total resistance in the circuit, including the internal resistance of the cell, is: \(R_{\text{total}} = R + r = 2\, \Omega + 1\, \Omega = 3\, \Omega\)
Using Ohm's law and the formula for terminal potential difference \(V\), which is given by: \(V = E - I \cdot r\) where \(I\) is the current through the circuit: \(I = \frac{E}{R_{\text{total}}} = \frac{3}{3} = 1\, A\)
Substitute the current back into the equation for terminal potential difference: \(V = 3 - (1 \times 1) = 3 - 1 = 2\, V\)
Therefore, the terminal potential difference of the cell is 2 V.
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Approach Solution -2
The circuit has a 3 V cell connected to resistances of \(1 \, \Omega\), \(4 \, \Omega\), and \(4 \, \Omega\). The total resistance \(R_{\text{total}}\) of the circuit is calculated as: \[ R_{\text{total}} = R_{\text{internal}} + R_{\text{external}} \] The external resistance is a parallel combination of \(4 \, \Omega\) and \(4 \, \Omega\): \[ R_{\text{parallel}} = \frac{1}{4} + \frac{1}{4} = 2 \, \Omega. \] Thus, the total resistance becomes: \[ R_{\text{total}} = 1 \, \Omega + 2 \, \Omega = 3 \, \Omega. \] The current in the circuit is: \[ i = \frac{\text{EMF}}{R_{\text{total}}} = \frac{3 \, \text{V}}{3 \, \Omega} = 1 \, \text{A}. \] The terminal potential difference \(V_{\text{terminal}}\) is given by: \[ V_{\text{terminal}} = \text{EMF} - i R_{\text{internal}} = 3 \, \text{V} - (1 \, \text{A} \cdot 1 \, \Omega) = 2 \, \text{V}. \] Final Answer: 2 V
