# Physics 5054 Doubts | Help Page 1

__Question 1: [Electricity and Magnetism > D.C. Circuits]__
R is a fixed resistor in the circuit.
Filament lamp is marked 12 V, 0.25 A.

Circuit is used to produce a
current/voltage graph for the filament lamp. Ammeter and voltmeter needed are
not shown.

To obtain different readings, the
student changes position of the movable contact.

**(a)**On Fig, draw the symbols for ammeter and voltmeter in the correct positions.

**(b)**Explain why it is sensible to include resistor R in this circuit.

**(c)**

(i) On Fig, sketch current/voltage
graph for the lamp.

(ii) State and explain how
current/voltage graph for a fixed resistor is different from graph for a
filament lamp.

**(d)**Fig shows position of the movable contact when the voltage across the lamp is 12 V and the current in the lamp is 0.25 A.

Determine

(i) current in the 50 Ω resistor

(ii) current in R

(iii) potential difference (p.d.)
across R

(iv) resistance of R

**Reference:**

*Past Exam Paper – June 2014 Paper 22 Q11*

__Solution 1:__**(a)**The ammeter and the voltmeter should be drawn with the correct symbols. The ammeter is in series with the lamp and the voltmeter is in parallel with the lamp.

**(b)**The resistor R limits or reduces the current / voltage. Otherwise the lamp blows

**OR**more of the 50 Ω can be used to adjust voltage / current.

**(c)**

(i) 12 V, 0.25 A correctly plotted (by
eye). The graph is a curved line from the origin with a correct curvature –
decreasing slope

(ii) The graph is a straight line (for
a fixed resistor). The

**lamp**has changing temperature or changing resistance**OR**the fixed resistor has constant temperature or constant**(d)**

(i) {From Kirchhoff’s law, the p.d. across the 50 Ω resistor is the same
as that across the filament lamp (since there are parallel in different loops),
which is stated to be 12V.}

(Current I =) V / R in any algebraic
or numerical for, e.g. 12/50

Current I = 0.24 A

(ii) {At a
junction, the current splits and go into the different components. So, the
current in the resistor R is the sum of currents passing through the components
in parallel.}

Current (= 0.25
+ 0.24) = 0.49 A

(iii) {The circuit
consists of a potential divider between resistor R and the parallel combination
of the lamp and the 50 Ω. From Kirchhoff’s law, the sum of p.d. across R and the
parallel combination should be equal to the e.m.f of 18V. Since the p.d. across
the parallel combination is 12V, the p.d. across R is 18 – 12 = 6V.}

p.d. = 6(.0) V

(iv) {The
current through R is 0.49A and the p.d. across it is 6V. From Ohm’s law, R = V
/ I = 6 / 0.49 = 12.24Ω.}

Resistance = 12(.24) Ω

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