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Question
A battery of emf 12 V and internal resistance 2 Ω is connected with two resistors A and B of resistance 4 Ω and 6 Ω respectively joined in series.
Find:
1) Current in the circuit
2) The terminal voltage of the cell
3) The potential difference across 6Ω Resistor
4) Electrical energy spent per minute in the 4Ω resistor.
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Solution
E = 12 V; ri = 2 Ω; RA = 4 Ω; RB = 6 Ω
1) The current in the circuit is
`I = E/R_"total" = E/(r_i + r_A + r_B)`
`:. I = 12/(2+4+6) = 1 A`
2) The terminal voltage of the cell is
V = E - Iri
∴ V = 12 - (1 x 2) = 12 - 2 = 10 V
3) The potential difference across the 6 Ω is
VB = IRB
∴ VB = 1 x 6 = 6V
4) The electrical energy spent per minute (=60 s) is
E = I2Rt
E = 12 X 4 X 60 = 240J
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