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**Question 34:**

(a) With the help of a circuit diagram, obtain the relation for the equivalent resistance of two resistances connected in parallel.

In the circuit diagram shown below, find :

(i) Total resistance.

(ii) Current shown by the ammeter A

**Solution :**

Suppose total current flowing the circuit is I. then the current passing through resistance R_{1} wiII be I_{1} and current passing through resistance R_{2} will be l_{2} .

Total current = I = I_{1}+I_{2}

Let resultant resistance of this parallel combination is R. By applying the Ohm's law to the whole circuit. we get that I = V/R

Since the potential difference across the both the resistances is same. so applying the Ohm's law to each resistance we get that

I_{1} = V/R_{1}

I_{2 }= V/R_{2}

Putting these eq in the above one, we get that

V/R = V/R_{1}+V/R_{2}

1/R = 1/R_{1}+1/R_{2}

If two resistance are connected in parallel than. the resultant resistance will be

1/R = 1/R_{1}+1/R_{2}

(b)

(i)Total reisitance = R

1/R=1/R_{1}+1/R_{2}

R_{2 }= 3+2=5ohms

R_{1} = 5ohms

1/R=1/5+1/5

1/R=2/5

R=2.5ohms

Ili) Current flowing through the circuit

I = V/R = 4/(2.5)

= 1.6amps

**Question 35:****(a) Explain with the help of a labelled circuit diagram, how you will find the resistance of a combination of three resistors of resistances R _{1}, R_{2} and R_{3} joined in parallel.**

(b) In the diagram shown below, the cell and the ammeter both have negligible resistance. The resistors are identical.

**With the switch K open, the ammeter reads 0.6 A. What will be the ammeter reading when the switch is closed ?****Solution :**

Suppose total current flowing in the circuit is I. then the current passing through resistance R_{1} will be I_{1} Lcurrent passing through resistance R_{2} will be I2 and current passing through resistance R_{3 }will bell I_{3}.

Total current = 1 = I_{1}+I_{2}+I_{3}

Let resultant resistance of this parallel combination is R. By applying the Ohm's law to the whole circuit. we get that

I=V/R

Since the potential difference across all the resistances is same. so applying the Ohm's law to each resistance we get that

l_{1} = V/R_{1}

I_{2} = V/R_{2}

I_{3} = V/R_{3}

Putting these eqs. in the above one. we get

V/R = V/R_{1}V/R_{2}+ V/R_{3} .

1/R = 1/R_{1} + 1/R_{2}+1/R_{3}

If two resistance are connected in parallel, then the resultant resistance will be

1/R = 1/R_{1}+1/R_{2}+1/R_{3}

(b) If switch is open. then only upper two resistances (connected in parallel) are in the circuit

Effective resistance is 1/R_{eq} = 1/R+1/R = 2/R

R_{eq} = R/2

So the current.I = VA(R/2) =0.6A (given)

V/R :0.3A

When the switch closes, the third resistance also comes in the circuit. The effective resistance of the circuit becomes R/3

Hence. Current 1= V/(R/3) = 3 (V/R) = 3 x 0.3 .0.9 A

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