Home
Class 12
PHYSICS
For the circuit shown, with R(1)=1.0Omeg...

For the circuit shown, with `R_(1)=1.0Omega,R_(2)=2.0Omega,E_(1)=2V` and `E_(2)=E_(3)=4V`, the potential difference between the points `'a'` and `'b'` is approximately (in `V`):

Text Solution

Verified by Experts

The correct Answer is:
3.33
Let us assume `V_(b) = 0V and V_(a) = xV`
`i_(1) + i_(2) + i_(3) = 0`
`implies (x - E_1)/(2R_1) + (x = E_2)/(R_2) + (x - E_3)/(2R_1) = 0`
`=(x - 2)/2 + (x - 4)/(2) + (x - 4)/2 = 0`
`implies 3x - 10 = 0 implies x = 3.3 V`
Promotional Banner

Topper's Solved these Questions

  • JEE MAIN REVISION TEST - 10| JEE -2020

    VMC MODULES ENGLISH|Exercise PHYSICS (SECTION 2)|5 Videos

  • JEE MAIN REVISION TEST - 1 | JEE - 2020

    VMC MODULES ENGLISH|Exercise PHYSICS ( SECTION 2)|5 Videos

  • JEE MAIN REVISION TEST - 12

    VMC MODULES ENGLISH|Exercise PHYSICS|25 Videos

Similar Questions

Explore conceptually related problems

In the given network of ideal cells and resistors, R_(1) = 1.0 Omega, R_(2) = 2.0 Omega, E_(1) = 2V and E_(2) = E_(3) = 4V . The potential difference between the point a and b is

In the circuit shown in fig. E_(1)=3V, E_(2)= 2V, E_(3)= 1V, R=r_(1)=r_(2)=r_(3)= 1 ohm . Find the potential difference between the points A and B and the currents through each branch.

Potential difference between points A and B (i.e. V_(A) - V_(B) ) is -----

In the circuit shown in figure , (epsilon)_(1)=3V,(epsilon)_(2)=2V,(epsilon)_(3)=1V and r_(1)=r_(2)=r_(3)=1(Omega) .Find the potential difference between the points A and B and the current through each branch.

In the circuit shown in fig. E_(1)=3" volt",E_(2)=2" volt",E_(3)=1" vole and "R=r_(1)=r_(2)=r_(3)=1 ohm. (i) Find potential difference in Volt between the points A and B with A & B unconnected. (ii) If r_(2) is short circuited and the point A is coneted to point B through a zero resistance wire, find the current through R in ampere.

In the circuit in figure E_1=3V, E_2=2V, E_3=1V and R=r_1=r_2=r_3=1Omega Find the potential differece between the points A and B and the currents through each branch.

In the circuit shown here, E_(1) = E_(2) = E_(3) = 2 V and R_(1) = R_(2) = 4 ohms . The current flowing between point A and B through battery E_(2) is

In the given circuit, R_(1) = 10Omega R_(2) = 6Omega and E = 10V, then correct statement is

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

In the circuit shown in figure E_1=10 V, E_2=4V, r_1=r_2=1Omega and R=2Omega . Find the potential difference across battery 1 and battery 2.