Question
1. Before being connected in a circuit, the emf of a certain dry cell is measured and found to be 4.5 V . When it is connected to the circuit, it supplies a current of 0.75 A , but the voltage reading across the dry cells terminals read 4.25 V . Using this information determine; (a) The voltage drop across the dry cells (b) The internal resistance of the dry cell when it is supplying current in this circuit.
Ask by Nguyen Barnett. in Papua New Guinea
Feb 28,2025
Upstudy AI Solution
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Answer
(a) The voltage drop across the dry cell is 0.25 V.
(b) The internal resistance of the dry cell is approximately 0.33 Ω.
Solution
Given:
• No-load emf (E) = 4.5 V
• When connected, current (I) = 0.75 A
• Terminal voltage (V_terminal) = 4.25 V
Step 1. Determine the voltage drop (V_drop) across the internal resistance:
V_drop = E − V_terminal = 4.5 V − 4.25 V = 0.25 V
Step 2. Determine the internal resistance (r) using Ohm’s Law:
r = V_drop / I = 0.25 V / 0.75 A ≈ 0.33 Ω
Final Answers:
(a) The voltage drop across the dry cell is 0.25 V.
(b) The internal resistance of the dry cell is approximately 0.33 ohms.
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The voltage drop across the dry cell can be calculated by finding the difference between the emf and the terminal voltage. So, it's 4.5 V - 4.25 V = 0.25 V. To find the internal resistance of the dry cell, we use Ohm's Law (V = IR). Rearranging gives us resistance (r) = V/I. The internal resistance, therefore, is 0.25 V / 0.75 A = 0.33 ohms. Now you know the voltage drop is 0.25 V and the internal resistance is about 0.33 ohms!
