Question

(a) Explain with the help of a labelled circuit diagram, how you will find the resistance of a combination of three resistors of resistances R1, R2 and R3 joined in parallel.
(b) In the diagram shown below, the cell and the ammeter both have negligible resistance. The resistor are identical.
Answer
(a)
Let the resistance of the three resistors be R1, R2 and R3, respectively. Let their combined resistance be R. Let the total current flowing in the circuit be I and the strength of the battery be V. Then from Ohm's law, we have
V = IR .....(1)
We know that when the resistors are connected in parallel, the potential drop across each resistance is the same.
Therefore
I = I1 + I2 + I3
I = V/R1 + V/R2 + V/R3
I = V/(1/R1 + 1/R2 + 1/R3) ...... (2)
From equations (1) and (2) we have
1/R = 1/R1 + 1/R2 + 1/R3
(b) Let V be the voltage of the cell and R is the resistance of each resistor. When switch K is not closed then
VR2=0.6 .....(1) [Using V = IR]Let A be the current, when switch (K) is closed VR3 = A .....(2)Dividing equation (2) by (1), we getA0.6 = 1.5 A = 1.5×0.6 =0.9 A

Answer 1

The resistors are connected in parallel, then the potential difference across each resistor is the same.

Explanation:  

Let us consider R1, R2 and R3 be the resistances of the three resistors connected in parallel, respectively. So, their combined resistance is R.  

Now, let us consider that the total current that is flowing in the circuit be I and the strength/ potential difference of the battery is V.

From Ohm’s law, we get,  

V = IR ----- (1)          

Since the resistors are connected in parallel, then the potential difference across each resistor is the same.

So, total current, I = I1 + I2+ I3  

   $= \frac{V}{R1} } +\frac{V}{R2} } +\frac{V}{R3} } =\frac{V}{\frac{1}{R1 } +\frac{1}{R2} +\frac{1}{R3} } ....(2)$ ------ (2)

  $\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3}$ (Combined Resistance)

(b) Let V be the voltage of the cell and R is the resistance of each resistor. When switch K is open then,

VR2 = 0.6 ------ (1)

Using ohm’s law i.e., V = IR

Let the current be A

So, when the switch (K) is closed  

VR3 = A ------ (2)

Dividing equation (2) by (1), we get,

$A 0.6 = 1.5 A = 1.5\times0.6 =0.9 A.$

You can get all the answers of Chapter 1 in the link below:

https://brainly.in/question/14961133

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Answer 2

Let the resistance of the three resistors be R1, R2 and R3, respectively. Let their combined resistance be R. Let the total current flowing in the circuit be I and the strength of the battery be V. Then from Ohm's law, we have

V = IR ..(1)

We know that when the resistors are connected in parallel, the potential drop across each resistance is the same.

Therefore

I = I1 + I2 + I3

I = V/R1 + V/R2 + V/R3

I = V/(1/R1 + 1/R2 + 1/R3) ... (2)

From equations (1) and (2) we have

1/R = 1/R1 + 1/R2 + 1/R3

(b) Let V be the voltage of the cell and R is the resistance of each resistor. When switch K is not closed then

VR2=0.6 ..(1) [Using V = IR]Let A be the current, when switch (K) is closed VR3 = A ..(2)Dividing equation (2) by (1), we getA0.6 = 1.5 A = 1.5×0.6 =0.9 A