1. In the circuit shown in Figure, an ideal ohmmeter is connected across ab with
ID: 1421065 • Letter: 1
Question
1. In the circuit shown in Figure, an ideal ohmmeter is connected across ab with the switch S open. All the connecting leads have negligible resistance. What will be the ohmmeter reading in Ohms (with two decial places) if R1 = 8.8 ?; R2 = 7.8 ? and R3 = 7.4 ??
2. Find the resistance of a 23 meter length of metal wire of 0.4 mm diameter. The resistivity of metal is 1.7×10-6 ?·cm at a temperature of 20°C. Express the answer with two decimal places.
3. Calculate the power dissipated in the 2 ? resistor, if the voltage of the battarey is ?=27.2 V and the resistance of the resistor is R=4.4 Ohms? Express the answer to two decimal places.
10 15 R2 R1 R3 20Explanation / Answer
1. A swe know, 1/Rtotal=1/(1/R1+1/R2+1/R3)
1/Rtotal= 1/((1/8.8)+(1/7.8)+(1/7.4))
1/Rtotal =1/0.37698= 2.65ohms(approx to 2 decimals)
As given in the circuit (10+20+15)= 45 ohms
This is parallel with 2.65ohms.
Equivalent resistance = (2.65 x 45)/(2.65 + 45)
= 2.50 ohms measured across AB.
2.As Resistivity is in ·cm, so we will work in cm units.
Length L = 23m = 2300cm
Radius of wire is r = D/2 = 2.0*10^-2 cm
Cross-sectional area A = pi*r^2 = 1.257*10^-3 cm^2
Resistance = resistivity * L / A
= 3*10^-6 * 2300 / 1.257*10^-3
= 5.49 ohms
Therefore, Resistance = 5.49 ohms
3. As we know, Power = Voltage^2/R
p = V^2/R
As given voltage = 27.2
Since,Circuit is parallel and sereis combination of resistors, Resistance = (4.4 + 2) = 6.4 ohms
p = 27.2^2/6.4
= 739.84/6.4
Therefore, Power = 115.6 Watts
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