Assume that each diode as shown in the figure has a forward bias resistance of 50 Ω and an infinite reverse bias resistance. The current through the resistance 150 Ω is
0.66 A
0.05 A
Zero
0.04 A
Two cells A and B of emf 2V and 1.5 V respectively, are connected as shown in figure through an external resistance 10 Ω. The internal resistance of each cell is 5 Ω. The potential difference EA and EB across the terminals of the cells A and B respectively are
EA = 2.0 V, EB = 1.5 V
EA = 2.125 V, EB = 1.375 V
EA = 1.875 V, EB = 1.625 V
EA = 1.875 V, EB = 1.375 V
C.
EA = 1.875 V, EB = 1.625 V
The figure can be redrawn as
The current through the circuit
The terminal potential difference of the batteries
VA = εA − irA = 2 − 0.025 × 5
= 2 − 0.0125 = 1.875 V
and VB = εB + irB = 1.5 + 0.025 × 5
= 1.5 + 0.0125
= 1.625 V