Senior Physics - Solutions
## Chapter 22-Electric Circuits- Worked Solutions

### ...to selected Complex reasoning process questions from the OUP text :-

*"New Century" Senior Physics - Knowledge, Processes and Reasoning*
by Walding, Rapkins and Rossiter.

**Q10 :-**
Given EMF = 12 V and source internal resistance r = 0.4W, the total circuit resistance

R_{tot} = 15 + 15 + 0.4 = 30.4W

Circuit current calculated from I = EMF/R_{tot} = 12/30.4 = 3.9 A.

Voltmeter will read 12 - (3.9 x 0.4) = 10.4 V and the ammeter will read 3.9 A as the circuit current.

**Q35 :-**
Each headlight dissipates 60 W at 12 V = 120 W, while each parking light dissipates 5 W at 12 V = 20 W. Thus total dissipation = 140 W. As the battery is rated at 60 AH and the total current drawn I = P/V = 11.7 A

Battery will last (60/11.7) = 5.1 hours.

**Q36 :-**
Section CD has R = 1.8W. and V_{A} = V_{B} for zero current. Using proportional division of voltage we can equate 0.2/5 = (0.8 x 5)/R which yields R = (0.8 x 5)/0.2 = 20.0W.

**Q37 :-**
(a) Zero voltage difference between the opposite ends of the bridge containing the switch and galvanometer.

(b) R1-R2 and R3-Rx divide supply voltage V into equal parts, hence

(R1 x V)/(R1 + R2) = (R3 x V)/(R2 + R3)

or R1/(R1 + R2) = R3/(R2 + Rx)

or R1.R3 + R1.Rx = R1.R3 + R3.R2 which rearranges to Rx = R3.R2/R1 as required.

(c) If R1 = 710W R2 = 317W R3 = 2.24W, then Rx by substitution is 1.0 kW.

**Q38 :-**
The equivalent circuit needs to be drawn with parallel(//) and series (+) sections. When this is done the circuit reduces to a simpler form as :-

(a) The total circuit resistance R_{tot} = 40 + 40 + (60//(5 + 15//(10 + 20)))W.

R_{tot} = 40 + 40 + (60//5 + 10) = 80 + 12 = 92W.

(b) Voltage drop across V_{(10+20)} = 1.0A x 30W = 30 V

hence current through 15W resistor must = 2.0A as it also has 30V across it.

while current through resistor 5W resistor = 1.0 + 2.0 = 3.0A from junction law.

which means that voltage drop across 5W resistor = 3.0 x 5 = 15 V

This allows calculation of a total voltage across 60W resistor of 15 + 30 = 45 V.

**Q39 :-**
Let the circuit current be (I) and the voltage across R2 be (V).

Hence EMF = 120 V = 100.I + V and also V.I = 30 as given from power dissipation. thus solving simultaneous equations created :-

120 = 100.I + 30/I which reduces to 10.I^{2} - 12.I + 3 = 0

solving the quadratic yields I = 0.84 A or I = 0.36 A as dual solutions.

Hence using R =P/I^{2} gives R = 232 or 43W.

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