You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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.

Open in App

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.

Suggest Corrections

Similar questions

Q. A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of

38 ω

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

Q. A

5 V

battery of negligible internal resistance is connected across a

200 V

battery and a resistance of

39 Ω

as shown in the figure. Find the value of the current.

Q. A solenoid of inductance

L

with resistance

r

is connected in parallel to a resistance

R

. A battery of emf

E

and of negligible internal resistance is connected across the parallel combination as shown in the figure. At the time

t = 0

, switch

S

is opened, calculate current through the solenoid after the switch is opened.

Q. A solenoid of inductance

L

with resistance

r

is connected in parallel to a resistance

R

. A battery of emf

E

and of negligible internal resistance is connected across the parallel combination as shown in the figure. At the time

t = 0

, switch

S

is opened, calculate amount of heat generated in the solenoid.

In the circuit shown in figure, a 12 V power supply with unknown internal resistance $r$ is connected to a battery with unknown emf $E$ and internal resistance 1 $Ω$ and to a resistance of 3 $Ω$ carrying a current of $2 A$ . The current through the rechargeable battery is $1 A$ in the direction shown. Find the unknown current $i$ , internal resistance $r$ and the emf $E$ .

Join BYJU'S Learning Program

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.