In the circuits shown above, a battery with a voltage ?V bat = 15 V is connected
ID: 2301187 • Letter: I
Question
In the circuits shown above, a battery with a voltage ?Vbat = 15 V is connected to the circuit with resistances R1 = R2 = 40 ohms, unknown resistor R3, and a switch connected to R3.
The volt meter measures the potential difference between the points A and B, ?VAB.
Define ?VO be the reading on the voltmeter when the switch is open and ?VC be the reading on the voltmeter when the switch is closed.
If the ratio (?VO / ?VC) = 0.80, what is the value of the resistance of R3?
Give your answer in ohms to three signficant digits. Do not include units in your answer.
In the circuit shown above, a battery with a voltage ?Vbat = 21 V is connected to the circuit with resistances R1 = 5 Ohms, R2 = 15 Ohms, and R3 = 40 Ohms.
The volt meter measures the potential difference between the points A and B, ?VAB.
What is the power lost in the resistor R3?
Give your answer in Watts to three signficant digits. Do not include units in your answer.
Explanation / Answer
a) Something is wrong in question as V0 will always be greater then Vc so ratio is wrong
I have used its reciprocal in my calculations
equivalent resistance when key open R = R1 + R2 = 40+40 = 80ohms
current = Vbat/R = 15/80 = 0.1875 amps
Vab = Vo = voltage across R2 = 0.1875*40 = 7.5V
Vc/Vo = 0.80 so Vc = 7.5*0.8 = 6 V
So voltage across R1 = 15-6 = 9 V
So current = 9/40 = 0.225 A
equivalent resistance of R2||R3 = 6/0.225= 26.67ohms
so 1/26.67 = 1/40 + 1/R3
so R3 = 80.03 ohms
b) equivalent resistance = 5+15||40 = 15.91ohm
current = 21/15.91 = 1.57 A
Vab = 21-(1.57*5) = 13.14V
so power lost in R3 = 13.142/40 = 4.32 watts