A cell of e.m.f. 2 V and internal resistance 1.2 Ω is connected to an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in figure.
Find (a) the reading of the ammeter, (b) the potential difference across the terminals of the cell, and (c) the potential difference across the 4.5 Ω resistor.

Question

A cell of e.m.f. 2 V and internal resistance 1.2 Ω is connected to an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in figure.

Find (a) the reading of the ammeter, (b) the potential difference across the terminals of the cell, and (c) the potential difference across the 4.5 Ω resistor.

KnowledgeBoat · Accepted Answer

Given, e.m.f. = 2V I = ? In the circuit, there are two parts. In the first part, resistors of 4.5 Ω and 9 Ω are connected in parallel. If the equivalent resistance of this part is R_p then In the second part, 1.2 Ω, 0.8 Ω and R_p = 3 Ω are connected in series. If the equivalent resistance of this part is R_s then Hence, the effective resistance of the circuit = 5 Ω Using Ohm's law, V = IR Hence, the reading of the ammeter = 0.4 A (b) The potential difference across the ends of the cells = ? ε = 2 V I = 0.4 A r = 1.2 Ω From relation, Voltage (V) = ε – Ir Substituting the values in the formula we get, Hence, potential difference across the terminals of the cell = 1.52 V (c) The potential difference across the 4.5 Ω resistor = ? Current flowing is I = 0.4 A. Now the current I divides in two parts. Let the current in 4.5 Ω resistor be I₁ and in 9 Ω resistor be I₂. So I = I₁ + I₂ and I₁ x 4.5 = I₂ x 9 On solving, p.d. across the 4.5 Ω resistor Alternate Method: V_4.5Ω = V_cell - V_ammeter Hence, p.d. across the 4.5 Ω resistor = 1.2 V

Physics

Current Electricity

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