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प्रश्न
A cell of e.m.f 2.0 V and internal resistance 1Ω is connected to the resistors of 3Ω and 6Ω in series. Calculate:
(i) the current drawn from the cell,
(ii) the p.d. across each resistor,
(iii) the terminal voltage of the cell and
(iv) the voltage drop.
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उत्तर
Total resistance of the circuit R = 1 + 3 + 6 = 10 Ω
(i) The current drawn from the cell = `"E"/"R" = 2.0/10 = 0.2` A
(ii) The p.d. across 3 Ω resistor V1 = IR1 = 0.2 × 3 = 0.6V
The p.d. across 6 Ω resistor V2 = IR2 = 0.2 × 6 = 1.2 V
(iii) The terminal voltage of the cell V = V1 + V2
= 0.6 + 1.2 = 1.8 V
(iv) The voltage drop = E - V = 2.0 - 1.8 = 0.2 V
संबंधित प्रश्न
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.
Potential difference is measured in _________ by using a ___________ placed in ___________ across a component.
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Write an expression for the electrical power spent in flow of current through a conductor in terms of resistance and potential difference.
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