Home
Class 12
PHYSICS
Figure shows a circuit with three ideal ...

Figure shows a circuit with three ideal batteries in it. The circuit elements h ave the following values

`DeltaV_(B_(1))=3.0V,DeltaV_(B_(2))=6.0V`
`R_(1)=2.0Omega,R_(2)=4.0Omega` The currents `i_(1) ,i_(2)` and `i_(3)` as shown in the circuit have the values.

A

0.50A,-0.25A,+0.25A

B

0.25A,-0.50A,-0.25A

C

0.50A,0.50A,1.0A

D

`-0.25A,0.50A,0.25A`

Text Solution

Verified by Experts

The correct Answer is:
1

Applying kirchhoff's first law at junction E, we g-t
`i_(3)=i_(1)+i_(2)`
Applying kirchhoff's second law for the closed loop
ABEFA, we get
`-i_(1)R_(1)+DeltaV_(B_(2))+i_(2)R_(2)-i_(1)R_(1)-DeltaV_(B_(1))=0`
`-2i_(1)R_(1)+i_(2)R_(2)+DeltaV_(B_(2))=DeltaV_(b_(1))=0`
`-2i_(1)(2)i_(2)(4)+6-3=0`
`-4i_(1)+4i_(2)+3=0`
`-4i_(1)+4i_(2)=-3`
Again, applying kirchhoff's second for the closed loop BCDEB, we get
`i_(3)R_(1)+DeltaV_(B_(2))-i_(3)R_(1)-i_(2)R_(2)-DeltaV_(B_(2))=0`
`-2i_(3)R_(1)- i_(2)R_(2)=0`
`-2i_(3)(2)-i_(2)(4)=0`
`ri_(3)+4i_(2)=i_(3)+i_(2)=0`
Solving (i),(ii) and (iii), we get
`i_(1)=0.50AA,i_(2)=-0.25A,i_(3)=0.25A`
Promotional Banner

Similar Questions

Explore conceptually related problems

In the following circuit , E_1 = 4V, R_1 = 2 Omega, E_2 = 6 V, R_2 = 2Omega, and R_3 = 4Omega. Find the currents i_1 and i_2

In the circuit shown, the currents i_(1) and i_(2) are

In the circuit shown below E_(1) = 4.0 V, R_(1) = 2 Omega, E_(2) = 6.0 V, R_(2) = 4 Omega and R_(3) = 2 Omega . The current I_(1) is

The current I_(1) (in A) flowing through 3Omega resistor in the following circuit is:

In the circuit shown in figure. epsi_(1)=3,epsi_(2)=2,epsi_(3)=6V,R_(1)=2,R_(4)=6Omega,R_(3)=2,R_(2)=4Omega and C=5muF . Find the current in the resistor R_(3) and the electrical energy stored in the capacitor C.

In the circuit shown, the value of currents I_(1), I_(2) and I_(3) are

For the circuit shown E=50 V R_(1)=10 Omega , R_(2)= 20 Omega, R_(3) = 30 Omega L = 2.0 mH . The current through R_(1) and R_(2) are ________ . immediate after the switch is closed.

In the circuit shown, E_(1)= 7 V, E_(2) = 7 V, R_(1) = R_(2) = 1Omega and R_(3) = 3 Omega respectively. The current through the resistance R_(3)

In the circuit shown, E_(1)= 7 V, E_(2) = 7 V, R_(1) = R_(2) = 1Omega and R_(3) = 3 Omega respectively. The current through the resistance R_(3)

In the circuit shown, the switch is closed at t=0 , the currents I_1, I_2 and I_3 are