# 1 Solutions G. DC Circuits G1 S1 and S2 closed, S3 ... - s3phytanwl

Electronics - Devices

Oct 7, 2013 (4 years and 9 months ago)

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1

Solutions

G.

DC Circuits

G1

S
1

and S
2

closed, S
3

opened. For this type of questions, just try drawing the circuit.

G2

Brightness of X does not change, since potential difference
across X (still in parallel with battery)
and resistance of X does not change.

Potential

difference across Y and Z are equal and half of that across X, since Y and Z are in
series. Therefore, Y and Z are equally bright and both dimmer than X.

G3

When temperature of surrounding increases, resistance of thermistor decreases. Therefore, since
thermistor is in series with a fixed resistor, ratio of resistance of thermistor to the resistance of the
fixed resistor decreases. Thus, potential difference a
cross the thermistor decreases.

P
otential difference
between the live

wire

(6

V)

and the neutra
l wire (0

V) is
held fixed by the
source.
Since voltmeter reads the potential difference across the live wire and ne
ut
ral wire,
the

Since resistance of thermistor decreases, total effective resistance of the circuit decreases.
Therefore, total current flowing through the circuits increases.

G4

When no light is on the LDR, ratio of resistance of LDR to

resistor R3 is 3600:1200. Therefore, the
potential difference across the LDR and R3 is in the ratio 3:1.

Potential difference across LDR = ¾ x 6.0 =
4.5 V
.

When light falls on the LDR, resistance of LDR is equal to the resistance of R3 .Therefore, the
potential difference across them are equal.

Potential difference across LDR =
3.0 V.

C
.

G5

(a)

Ohm's Law states that the current passing
through a metallic conductor is directly proportional to
the potential difference across its ends, provided the physical conditions (such as temperature)
are constant.

(b)

As potential difference across the lamp increases, current flowing through the lamp inc
reases at a
decreasing rate, as seen by the decreasing gradient of the graph. Therefore, the ratio of potential
difference is to current increases. Therefore, the resistance of the lamp increases.

(c)

At 3.0 A, potential difference across the lamp is about 4.5

V and potential difference across the
resistor is about 9.0 V. Therefore, potential difference across the battery is about 13.5 V.

G6

For this question, you can put all in and calculate and check if they are correct. Firstly, the top
resistor in the pa
rallel circuit has to have half the resistance of the one at the bottom. Based on
the options, for each one, we can calculate the potential difference across the parallel part of the
circuit. Whatever is left can be verified to see if they are coherent!

2

H.

Electromagnetic Induction

H1

(a)

As the magnet is pushed into the solenoid, the magnetic field within the solenoid changes. By
e.m.f. is induced in the coil.

(b)

Kinetic energy of the hand doing work against the opposing magnetic force due to the induced
current (Lenz's law) will be converted to electrical energy.

H2

(a)

As the turbine rotates, the magnetic field in the coil changes due to the motion of the magnets
on the turbine. According to Faraday's law, e.m.f. will be induced in the coil. A north pole will
be induced on the side of the coil nearer to the turbine to re
pel an approaching magnet and as
the magnet moves further away, a south pole will be induced on the same end, according to
Lenz's law. As the polarity changes, the direction of induced current
will flow in opposite
directions. Therefore, an alternating e.m
.f. is induced.

(b)

H
3

(a)

Turns

ratio of Transformer A

= 400/25 =
16

(b)

Turns

ratio of Transformer B

= 200/400000 =
5.0 x 10
-
4

(c)

When high voltage is used to transmit power, current in the transmission cables is reduced.
Therefore, less power will be lost
as heat in the cables.

(d)

It is given that the total resistance of the transmission cables is 1000

. If the electric power
produced at the power station is 4000 kW, and Transformer A is 90% efficient, calculate the
power loss in the cable.

Power

dissipated by secondary coil of transformer A at the start of the transmission cables

= 0.90 x 4000 kW

= 3600 kW

Therefore, current in transmission cables = 3600 kW / 400 kV = 9.0 A

Power loss in cable = (9.0)
2

. 1000

= 81 000 W

=
81 kW

H3

No, the
coil will not continue rotating as e.m.f. will be induced in the coil as the coil cuts magnetic
field lines. Since coil is connected to a closed circuit, current will flow. According to Lenz's law, the
direction of induced current will be so as to produce
a magnetic effect that opposes the motion of
the coil. Therefore, the force acting on the coil due to the induced current will always oppose the
motion of the coil, resulting in the coil eventually slowing down

to a stop
.