Electricity

Activity . Make a parallel combination, XY, of three resistors having resistances R , R , and R , respectively. Connect it with a battery, a plug key and an ammeter, as shown in Fig. .

. Also connect a voltmeter in parallel with the combination of resistors. Plug the key and note the ammeter reading. Let the current be I .

Also take the voltmeter reading. It gives the potential difference V, across the combination. The potential difference across each resistor is also V . This can be checked by connecting the voltmeter across each individual resistor (see Fig.

. ). Figure . Figure .

Figure . Figure . Figure . Take out the plug from the key.

Remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor R , as shown in Fig. . .

Note the ammeter reading, I . Figure . Figure . Figure .

Figure . Figure . Similarly, measure the currents through R and R . Let these be I and I , respectively.

What is the relationship between I , I , I and I ? It is observed that the total current I , is equal to the sum of the separate currents through each branch of the combination. I = I + I + I ( . ) Let R p be the equivalent resistance of the parallel combination of resistors.

By applying Ohm’s law to the parallel combination of resistors, we have I = V/R p ( . ) On applying Ohm’s law to each resistor, we have I = V /R ; I = V /R ; and I = V /R ( . ) From Eqs. ( .

) to ( . ), we have V/R p = V/R + V/R + V/R or /R p = /R + /R + /R ( . ) Thus, we may conclude that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances. Example .

In the circuit diagram given in Fig. . , suppose the resistors R , R and R have the values Ω , Ω , Ω , respectively, which have been connected to a battery of V. Calculate (a) the current through each resistor, (b) the total current in the circuit, and (c) the total circuit resistance.

R = Ω , R = Ω , and R = Ω . Potential difference across the battery, V = V. This is also the potential difference across each of the individual resistor; therefore, to calculate the current in the resistors, we use Ohm’s law. The current I , through R = V/ R I = V/ Ω = .

A. The current I , through R = V/ R I = V/ Ω = . A. The current I , through R = V/R I = V/ Ω = .

A. The total current in the circuit, I = I + I + I = ( . + . + .

) A = A The total resistance R p , is given by [Eq. ( . )] p + + Thus, R p = Ω . Example .

If in Fig. . , R = Ω , R = Ω , R = Ω , R = Ω , R = Ω , and a V battery is connected to the arrangement. Calculate (a) the total resistance in the circuit, and (b) the total current flowing in the circuit.

Suppose we replace the parallel resistors R and R by an equivalent resistor of resistance, R ′ . Similarly we replace the parallel resistors R , R and R by an equivalent single resistor of resistance R ″ . Then using Eq. ( .

), we have / R ′ = / + / = / ; that is R ′ = Ω . Similarly, / R ″ = / + / + / = / ; that is, R ″ = Ω . Thus, the total resistance, R = R ′ + R ″ = Ω. To calculate the current, we use Ohm’s law, and get I = V/R = V/ Ω = .

A. We have seen that in a series circuit the current is constant throughout the electric circuit. Thus it is obviously impracticable to connect an electric bulb and an electric heater in series, because they need currents of widely different values to operate properly (see Example . ).

Another major disadvantage of a series circuit is that when one component fails the circuit is broken and none of the components works. If you have used ‘fairy lights’ to decorate buildings on festivals, on marriage celebrations etc., you might have seen the electrician spending lot of time in trouble-locating and replacing the ‘dead’ bulb – each has to be tested to find which has fused or gone. On the other hand, a parallel circuit divides the current through the electrical gadgets. The total resistance in a parallel circuit is decreased as per Eq.

( . ). This is helpful particularly when each gadget has different resistance and requires different current to operate properly. Figure .

Figure . Figure . Figure . Figure .

An electric circuit showing the combination of series and parallel resistors