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
Rtot = 15 + 15 + 0.4 = 30.4W
Circuit current calculated from I = EMF/Rtot = 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 VA = VB 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 Rtot = 40 + 40 + (60//(5 + 15//(10 + 20)))W.
Rtot = 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.I2 - 12.I + 3 = 0
solving the quadratic yields I = 0.84 A or I = 0.36 A as dual solutions.
Hence using R =P/I2 gives R = 232 or 43W.