Home
Class 12
PHYSICS
Two resistors are connected in series wi...

Two resistors are connected in series with 5V battery of negligible internal resistance. A current of 2 A flows through each resistor. If they are connected in parallel with the same battery a current of `(25)/(3)A` flows through combination. Calculate the value of each resistance.

Text Solution

Verified by Experts

Resistors in series `: R _(S) = R_(1) + R _(2)`
`I = (epsi )/(R _(s))`
`R _(S) = (epsi )/(I) =5/2`
`R_(1) + R _(2) =2.5`
Resistors in parallel `R _(p) = (R _(1) R_(2))/( R _(1) + R _(2))`
`I = (epsi )/(R _(p))`
`R _(p) = (epsi )/(I) = (5)/( 25//3) = 3/5`
`R_(1) R _(2) = 3/5 xx (R_(1) + R _(2)) = 3/5 xx 2.5`
`R_(1) R _(2) = 1.5`
Solving equations (1) and (2),` R_(1) = 1.5 Omega and R_(2) = 1Omega `
Promotional Banner

Topper's Solved these Questions

  • II PUC ANNUAL EXAMINATION 2019

    OSWAAL PUBLICATION|Exercise PART-C|8 Videos

  • ELECTROSTATIC POTENTIAL AND CAPACITANCE

    OSWAAL PUBLICATION|Exercise TOPIC -2 CAPACITORS AND CAPACITANCE (NUMERICAL PROBLEMS)|8 Videos

  • II PUC MARCH-2017 (SOLVED PAPER )

    OSWAAL PUBLICATION|Exercise PART-D|11 Videos

Similar Questions

Explore conceptually related problems

Which two resistors are connected in series with a cell of emf 2V and negligible internal resistance, a current of (2/5)A flows in the circuit. When the resistances are in parallel, the main current is (5/3)A. Calculate the resistances.

If the combination is connected to a battery of emf 20 v and negligible internal resistance determine the current through each resistor , and the total current drawn from the battery .

A 4 V battery of internal resistance 1 Omega sends a current through a 3 Omega resistor. The terminal p.d. is

Two identical cells connected in series send 10 A current through a 5 Omega resistor. When they connected in parallel, they send 8 A current through the same resistance, the internal resistance of each cell is