Electricity Worksheet22

Two students perform experiments on series and parallel combinations of two given resistor R_{1} and R_{2} and plot the following V – I graphs,
Which of the graphs is (are) correctly labelled in terms of the words 'series' and 'parallel' ? Justify your answer.

You are given three resistors of 10 Ω , 10 Ω and 20 Ω; a battery of emf 2.5 V a key, an ammeter and a votmeter. Draw a circuit diagram showing the correct connections of all given components such that the voltmeter gives a reading of 2.0 V

Two electric circuits I and II are shown in the Fig. In circuit I, the key K is closed whereas in circuit II, the key is open. Compare the currents in the circuit I and circuit II.

An electric geyser has the rating 2000 W, 200 V marked on it What should be the minimum rating (in whole number) of a fuse wire that may be required for safe use with this geyser ?

An electrician puts a fuse of rating 5 A in that part of domestic electrical circuit in which electrical heater of rating 1.5 kW, 220 V is operating. What is likely to happen in this case and why ? What change, if any, needs to be made ?

The electric power consumed by a device may be calculated by using either of the two expressions P = I^{2}R or P = V^{2}/R. The first expression indicates that it is directly proportional to R whereas the second expression indicates inverse proportionality. How can this different dependence of P on R in these expressions be explained ?

A cylinder of a material is 10 cm long and has a crosssection of 2 cm^{2}. If its resistance along the length be 20 Ω, what will be its resistivity in number and units.

Why is tungsten metal selected for making filaments of incandescent lamps ?

A resistance of 10 ohm is bent in the form of a closed circle. What is the effective resistance between the two points at the ends of any diameter of this circle ?

Why is much less heat generated in long electric cables than in filaments of electric bulbs ?
Answer:

The resistance (R_{s}) in series combination of two resistors is more than the resistance (R_{p}) in parallel combination of the same two resistors, i.e., R_{s} > R_{p}. As I = V/R, for a given pd (V), I_{s} < I_{P}. In both the graphs, for a given pd (V) across each combination, I_{s} < I_{p}. Hence, both the graphs are properly labelled.

The circuit diagram showing the correct connections is shown in the fig. Since the two resistors, each of 10 Ω are in parallel combination, their equivalent resistance is given by
or R_{p} = 5 Ω
Since R_{p} and 20 Ω are in series combination, the total resistance in the circuit is given by
Current in the circuit,
pd across 20 Ω resistor

In circuit I, since K is closed, R_{1} and R_{2} are in parallel combination. Their equivalent resistance
Is given by
Current in the circuit,
In circuit II, since K is open, the resistance R_{2} is out of circuit and as such the net resistance in the circuit is R_{1}.
Current in the circuit,
Thus,
Since (R_{1} + R_{2}) > R_{2}, I_{1} > I_{2}.
Thus current in circuit I is more than the current in circuit II.

P = 2000 W, V = 200 V
As P = VI,
Thus, minimum current rating (in whole number) of the fuse wire = 10 A

P = 1.5kW= 1500W, V = 200 V.
P = VI,
When 6.81 A current flows through the fuse wire of rating 5A, it melts and the circuit breaks. Because the current flowing through the fuse, 6.81 A is more than its rating, i.e. 5 A.
In this case, a fuse of rating higher than 6.81 A need to be used.

Here both the expressions are correct and are used to calculate electric power. The expression P = I^{2}R is used to calculate electric power when the current flowing through the circuit is constant. This is the case when the electric devices are connected in series.
The expression P = V^{2}/R is used to calculate electric power when the pd (V) across the circuit is constant. This is the case when the electric devices are connected in parallel.

ohm cm.

Because it has high melting point.

Each of the two semicircles between the two ends of a diameter has a resistance of 10 Ω /2 = 5 Ω. These two resistance are in parallel between the two ends of the diameter. Thus, the effective resistance = 5 Ω/2 = 2.5 Ω

Electric cables are made of thick copper which have much less resistance as compared to thin tungsten filament of electric bulbs.