You are given a 48.5 V battery and two resistors of resistances 27 and 94.5 . (a

ID: 1549706 • 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.

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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