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Resistance are used to reduce the current ow in a circuit.

1. A carbon resistor has coloured strip and shown in the figure. What is its resistance?

2. Resistance can be connected in series and parallel to obtain the required value of resistance. Derive an expression for the effective resistance when three resistors are connected in parallel.

3. Kirchho’s rules are used to analyses the electric circuit. Use it to analyze the Wheatstone Bridge and arrive at Wheatstone’s condition for balancing the bridge.

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1. blue, Gray, yellow, gold

1st – Blue – 6

IInd – Gray – 8

IIIrd – Yellow – 104

Iv – gold – 5%

68 × 104 ± 5%.

2. Derive an expression for effective resistance in series: Consider three resistors R1 , R2 and R3 connected in series and a pd of V is applied across it.

In the circuit shown above the rate of ow of charge through each resistor will be same i.e. in series combination current through each resistor will be the same. However the pd across each resistor are different and can be obtained using ohms law.

pd across the first resistor V1 = I R1

pd across the second resistor V2 = IR2

pd across the third resistor V3 = I R3

If V is the effective potential drop and R is the effective resistance then effective pd across the combination is

V = IR 

Total pd across the combination = the sum pd across each resistor, V = V1 + V2 + V3

Substituting the values of pds we get IR = IR1 + IR2 + IR3

Eliminating I from all the terms on both sides we get

R = R1 + R2 + R3 ………(1) 

Thus the effective resistance of series combination of a number of resistors is equal to the sum of resistances of individual resistors.

3. Wheatstone’s Bridge: 

Four resistances P,Q,R and S are connected as shown in figure. Voltage ‘V’ is applied in between A and C. Let I1 , I2 , I3 and I4 be the four currents passing through P,R,Q and S respectively.

Working: 

The voltage across R

When key is closed, current flows in different branches as shown in figure. Under this situation 

The voltage across P, VAB = I1

The voltage across Q, VBC = I3 Q ……(1)

The voltage across R, VAD = I2 R

The voltage across S, VDC = I1

The value of R is adjusted to get zero deflection in galvanometer. Under this condition,

I1 = I3 and I2  = I4 ……(2)

Using Kircho’s second law in loop ABDA and BCDB, we get

VAB = VAD ……….(3) 

and VBC = VDC  …….. (4)

Substituting the values from eq(1) into (3) and (4), we get

I1 P = I2 R……….(5)

and I3Q = I4S……..(6)

Dividing Eq(5) by Eq(6)

\(\frac{I_1P}{I_3Q}=\frac{I_2R}{I_4S}\)

\(\frac{P}{Q}=\frac{R}{S}\) [since I1 = I3 and I2 = I4]

This is called Wheatstone condition.

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