Question

1. In class we derived how a transformer can up or downregulate the voltage in an electrical circuit with an alternating current.

(a) Use the conservation of energy to deduce how the currents I1 and I2 must be related if you know the ratio of ε1/ε2 (for the purpose of calculating the power delivered by a circuit ignore any complications due to the fact that I(t) and ε(t) are time-varying).

(b) In class I argued that power transmission is more efficient with high voltage lines compared to lower voltages. To illustrate this point let’s consider the following simple DC circuit. (In circuit diagrams the wires – by definition – do not have any resistance. If we want to calculate the real world losses of power lines we thus have to explicitly introduce a resistance due to the wires.) +_ V V

Rwire load

Consider the above circuit for ε = 101 V, Rwire = 1 Ω, Rload = 100 Ω. What is the power delivered by the battery to the circuit? What is the power lost across the wire? What is the power delivered to the load? (Hint: determine the overall resistance and thus the current and then calculate the voltage drop across each of the resistors.)

(c) Consider the above circuit for ε = 1000.1 V, Rwire = 1 Ω, Rload = 10, 000 Ω What is the power delivered by the battery to the circuit? What is the power lost across the wire? What is the power delivered to the load?

(d) What can you conclude from the above examples to illustrate that high voltage lines are more efficient in delivering electricity from a power plant to a city?

Answer 1

(a) By conservation of enegy, the input and output powers must be the same as no power is assumed to be lost. Therefore,

$\small \epsilon_{1}I_{1} = \epsilon_{2}I_{2} \\ \\ \frac{I_1}{I_2} = \frac{\epsilon_{2}}{\epsilon_{1}} = \frac{N_2}{N_1}$

where N's are the winding nos. of primary nd secondary circuits.

(b) Total resistance in the circuit = 101 ohms, therefore, current in the circuit, I = V/R = 101/101 = 1 ampere; total power delivered to the circuit by the battery is VI = 101 W

Power lost across the wire: $\small I^2 R = 1\times 1 = 1\ W$

Power delivered to the load: $\small I^2 R = 1\times 100 = 100$ W

(c) Total resistance = 10001 ohms, current in the circuit = 0.1 ampere; total power delivered = 100.01 W

Power lost across the wire = $\small I^2 R = 0.1^2 \times 1 = 0.01\; W$

Power delivered to the load = $\small I^2 R = 0.1^2 \times 10000 = 100\; W$

(d) In the second case, power lost across the wire is smaller despite wire resistance being the same, this is because power depends on current squared, which is smaller in the second case. The statement in the question is evidently true.

Answer 2

Hns a) Th conser vation of enerfr the transformer output = c2. e Lurrent in the Circuit Koad Rwo )