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Standard XII

Physics

Series and Parallel Combination of Resistors

A 10 V cell o...

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

Given: $E_{1} = 10 V; E_{2} = 200$ V, r = $38 Ω, I = ?$

Net emf, $E = E_{2} - E_{1}$ = 200 V - 10 V = 190 V

$∴ C u r r e n t, I = \frac{Net emf}{Resistance} = \frac{190}{38} = 5$ A.

Thus, a current of 5A flows in the circuit.

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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 38 Ω as shown in the figure. Find the value of current in the circuit.

Given: E1=10V;E2=200 V, r = 38 Ω,I=? Net emf, E=E2−E1 = 200 V - 10 V = 190 V ∴Current,I=Net emfResistance=19038=5 A. Thus, a current of 5A flows in the circuit.

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.

Given: $E_{1} = 10 V; E_{2} = 200$ V, r = $38 Ω, I = ?$

Net emf, $E = E_{2} - E_{1}$ = 200 V - 10 V = 190 V

$∴ C u r r e n t, I = \frac{Net emf}{Resistance} = \frac{190}{38} = 5$ A.

Thus, a current of 5A flows in the circuit.