Goodmarks
Sign in

Physics · Internal resistance, potential difference and emf of a cell, a combination of cells in series and parallel

A current of ampere flows through a resistance when connected across a cell of e

A current of \( I \) ampere flows through a resistance \( R \) when connected across a cell of emf \( E \) and internal resistance \( 1 \Omega \) When \( R \) is increased by \( 50 \% \), the current through the circuit is \( 0.8 A . \) The value of \( \boldsymbol{R} \) is :

  • A. \( 1 \Omega \)
  • B. \( 1.5 \Omega \)
  • C. 2 \Omega \)
  • D. 4\Omega

Step-by-step solution

Let initial current I = E/(R+1). When R increases by 50%, new resistance 1.5R, new current 0.8 = E/(1.5R+1). Eliminating E gives I(R+1) = 0.8(1.5R+1). Assuming I = 1 A (a common simple value), solving yields R = 1 Ω. Checking: if R=1, E = 2 V, initial I=1 A, new I=2/2.5=0.8 A, consistent.
Practise more in this unitView MCQsSign up for full question bank

Related MCQs