Problem 10: Two resistors are connected in series and then to a battery as shown

Question

Problem 10: Two resistors are connected in series and then to a battery as shown in the figure. a) If the emf of the battery is (E=1V), what is the current through each resistor (ignore the internal resistance r)? b) What is the voltage (V) across each resistor (VR1 and VE2)? c) The internal resistance (r) of the battery increases after the battery is used and as a result the current decreases. When the current through each resistor is dropped by 20% of the value you calculated in part (a) above, find: the emf (E) and the internal resistance (r) of the battery. d) What is the rate at which energy is dissipated in the internal resistance (power)? (use the value you found in part (c) for the internal resistance (r))

Explanation / Answer

a)i=1/5=0.2amp

b)vr1=2*0.2=0.4

vr2=3*0.2=0.6

c)i=0.2*0.8=0.16amp

0.16=1/(5+r)

r=1.25ohm

d)i2r=1.25*(0.16)2=0.032w

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