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Ohm's Law

Please only do review questions 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 2.10 and problems 2.2, 2.3, 2.4,2.5, and 2.6.

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SOLUTION

2.3 Power of the heater (i.e. the energy consumed by the heater per second) = P = VI = I2R = V2/R where V = Voltage applied, I = Current flowing, R = Resistance of the heater ...........(1)

Here, P = 1.5 kW = 1.5 x 1000 = 1500 W, I = 12 A.

Using P = VI and substituting values we get : 1500 = 12 x V or V = 125 Volts

2.4 Using P = I2R and substituting P = 2 W and R = 80 k ohm = 80000 ohm, we get :
________
2 = I2 x 80000 or I2 = 2/80000 = 25 x 10-6 or I = √25 x 10-6 = 5 x 10-3 A = 5 mA

I = 5 mA

2.5 Number of branches (b), number of independent loops (l) and number of nodes (n) are related by the equation : l = b - (n - 1).

Here, l = 8, b = 12. Substituting we get : 8 = 12 - (n - 1). Solving we get n = 5

2.6 As per Ohm,s law : I = V/R where I = Current in the circuit, V = Net voltage in the circuit and R = Total resistance in the circuit.
In the given circuit : R = 6 + 4 = 10 ohm
To find the net voltage we note that the two voltage sources tend to send currents in opposite directions. 3 V source sends current in clockwise direction and 5 V in anticlockwise ...

Solution Summary

This set of basic questions has been solved using Ohm's law, Kirchhoff's laws etc..

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