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Standard XII

Physics

Mixed Combination of Cells

A 10 V cell...

Question

A $10 V$ cell of negligible internal resistance is connected in parallel across a battery of emf $200 V$ and internal resistance $38 Ω$ as shown in the figure. Find the value of current in the circuit.

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Solution

Net emf of the combination of cell $E = 200 - 10 = 190 V$
Net resistance of the circuit $R = 38 Ω$
So, current in the circuit $I = \frac{E}{R}$
$⟹ I = \frac{190}{38} = 5 A$

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A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38Ω as shown in the figure. Find the value of current in the circuit.

Net emf of the combination of cell E=200−10=190 VNet resistance of the circuit R=38ΩSo, current in the circuit I=ER⟹ I=19038=5 A

A $10 V$ cell of negligible internal resistance is connected in parallel across a battery of emf $200 V$ and internal resistance $38 Ω$ as shown in the figure. Find the value of current in the circuit.

Net emf of the combination of cell $E = 200 - 10 = 190 V$
Net resistance of the circuit $R = 38 Ω$
So, current in the circuit $I = \frac{E}{R}$
$⟹ I = \frac{190}{38} = 5 A$