A cell of emf E and internal resistance r is connected across an external resistance R (R >> r). The p.d. across R is A 1
A. \(\frac {E}{R+r}\)
B. E(I – \(\frac {r}{R}\))
C. E(I + \(\frac {r}{R}\))
D. E (R + r)
Answer:B
A cell of emf E and internal resistance r is connected across an external resistance R (R >> r). The p.d. across R is A 1
A. \(\frac {E}{R+r}\) B. E(I – \(\frac {r}{R}\)) C. E(I + \(\frac {r}{R}\)) D. E (R + r)