MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Parallel Combination of Cells

In the circui...

Question

In the circuit shown below $E_{1} = 4.0 V, R_{1} = 2 Ω, E_{2} = 6.0 V, R_{2} = 4 Ω$ and $R_{3} = 2 Ω$ . The current $I_{1}$ is

A

1.6 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

1.8 A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

1.25 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

1.0 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 1.8 A
Applying Kirchhoff's law for the loops (1) and (2) as shown in figure For loop(1)

$- 2 i_{1} - 2 (i_{1} - i_{2}) + 4 = 0 \Rightarrow 2 i_{1} - i_{2} = 2 . . . . (i)$
For loop (2)
$- 2 (i_{1} - i_{2}) + 4 i_{2} - 6 = 0 \Rightarrow - i_{1} + 3 i_{2} = 3 . . . (i i)$
On solving equation (i) and (ii) $i_{1} = 1.8 A .$

Suggest Corrections

thumbs-up

0

Similar questions

Q. In the circuit shown

E_{1} = 4.0 V, R_{1} = 2 Ω, E_{2} = 6.0 V, R_{2} = 4 Ω

R_{3} = 2 Ω

I_{1}

Q. In the circuit shown below

E_{1} = 4.0 V, R_{1} = 2 Ω, E_{2} = 6.0 V, R_{2} = 4 Ω

R_{3} = 2 Ω .

l_{1}

Q. In the circuit shown below

E_{1} = 4.0 V, R_{1} = 2 Ω, E_{2} = 6.0 V, R_{2} = 4 Ω

R_{3} = 2 Ω

I_{1}

Q. In the following circuit

E_{1} = 4 V, R_{1} = 2 Ω

E_{2} = 6 V, R_{2} = 2 Ω

R_{3} = 4 Ω

i_{1}

Q. In the given circuit,

E_{1} = 3 V, E_{2} = 2 V, E_{3} = 6 V, R_{1} = 2 Ω, R_{4} = 6 Ω, R_{3} = 2 Ω, R_{2} = 4 Ω, C = 5 μ F

. The energy stored in the capacitor is :

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Cell and Cell Combinations

PHYSICS

Watch in App

explore_more_icon

Explore more

Parallel Combination of Cells

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon