Question

4. In the given circuit, calculate
i) net resistance of the circuit
ii) Voltmeter reading when
a) both the keys are open b) both the keys are closed iii) current through 3 Ω resistor.

Answer 1

THEORY :-

Series combination of resistance ( diagram (a) )

When resistors are arranged in such a way that the ending of the first resistor are connected to the starting point if the second resistor. (please see the diagram ( a) )

In such a case, equivalent resistance of a combination of resistances equals

Req = R1 + R2 + R3...........

• The current through any two resistors in series us always same.However, there may be voltage difference across them

parallel combination of resistance

When resistors are arranged in such a way that the are connected across the same point. ( see tha diagram b )

In such a case,

$\frac{1}{</strong><strong>Req</strong><strong>} = \frac{1}{</strong><strong>R</strong><strong>1</strong><strong>} + \frac{1}{</strong><strong>R2</strong><strong>}$

• The voltage difference across any two resistor in parallel is always same.

SOLUTIONS

We have ,

2 ohm and 3 ohm resistors are in series .

Req = 2 + 3 = 5.

Now, this will be parallel to 5 ohm resistor. Hence, we get two 5 ohm resistor in parallel.

Req = 5 /2.

(i) Hence, net resistance of the circuit is 5 ohm.

(ii) a . When key K1 will be closed, no current will pass through 5 ohm resistor , hence V = 5 I = 0. Hence, voltage difference will be 0.

(ii) b. When key K2 is closed, no current will pass through 2 and 3 ohm resistors, hence, all current will pass through 5 ohm.Hence, V across 5 ohm = 6 V = V across battery = reading of voltmeter.

(iii) Net current in the circuit = V/Req = 6/2.5 = 2.4

Now, this will distribute into two resistors of 5 ohm and 5 ohm.They both will get 2.4/2 = 1.2 A.

Now, 2 ohm and 3 ohm are in Series, Hence, current flowing through them will be equal and will be 1.2 A.

Answer 2

Explanation:

(1)Rs=3+2=5 ohm. Therefore 1/Rp=1/5+1/5=2/5

Therefore,Rp=5/2.=2.5 ohm.Therefore the total resistance of the circuit is 2.5ohm.(2)(a) when both the keys are open.Then the reading of ammeter and voltmeter is zero .(b)When key2 is open and key1 is closed then, total resistance of the circuit,R=5ohm. Given,V=10v, therefore reading of ammeter,I=V/R=10/5=2Ampere as well as reading of voltmeter is 10v.(c) When both the keys are closed,then total resistance =2.5 ohm,V=10v.Therefore, reading of ammeter=V/R=10/2.5=4Ampere and reading of voltmeter=10v.