You are given a 48.5 V battery and two resistors of resistances 27 and 94.5 . (a
ID: 1601331 • Letter: Y
Question
You are given a 48.5 V battery and two resistors of resistances 27 and 94.5 .
(a)Find the current in A when these resistors are connected in series with the battery.
(b) Find the power dissipated in W by the 27 resistor when connected in series with the rest of the circuit.
(c)Find the power dissipated in W by the 94.5 resistor when connected in series with the rest of the circuit.
(d) Find the current in A being drawn from the battery when the resistors are connected in parallel with the battery.
Explanation / Answer
part a:
when connected in series, net resistance in the circuit=27+94.5=121.5 ohms
current in the ressitors=total voltage/total resistance
=48.5/121.5=0.39918 A
part b:
power dissipated=current^2*resistance
=0.39918^2*27
=4.3023 W
part c:
power dissipated in 94.5 ohms resistor=94.5*0.39918^2=15.058 W
part d:
when two resistors are connected in parallel,
net resistance=27*94.5/(27+94.5)=21 ohms
net current=voltage/total resistance
=48.5/21=2.3095 A
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