Question

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 R1, R2 and R3 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 ?

Lakhmir Singh Physics Class 10

Answer 1

Answer:

R = R₁R₂R₃/ (R₁R₂ + R₂R₃ + R₁R₃)

Explanation:

V  = I₁R₁  => I₁ = V/R₁

V = I₂R₂  => I₂ = V/R₂

V = I₃R₃ => I₃ = V/R₃

I  = I₁ + I₂ + I₃

=> I  = V/R₁ +  V/R₂ + V/R₃

=> I = V ( 1/R₁  + 1/R₂   + 1/R₃)

=> 1/ ( 1/R₁  + 1/R₂   + 1/R₃)  = V/I

=> R = 1/ ( 1/R₁  + 1/R₂   + 1/R₃)

=> R = R₁R₂R₃/ (R₁R₂ + R₂R₃ + R₁R₃)

Answer 2

$\frac{1}{R_eff} }$ = $\frac{1}{R_{1} }$ + $\frac{1}{R_{2} }$ + $\frac{1}{R_{3} }$

$R_{eff}$ = R₁R₂R₃/ (R₁R₂ + R₂R₃ + R₁R₃)

Explanation:

When resistors are in parallel as shown in the attachment, the

potential difference across all the 3 resistors is the same and is equal to V.

According to Ohm's Law,   V = IR

∴ $I_{1}$ =  V/R₁  

   $I_{2}$ = V/R₂

   $I_{3}$ =  V/R₃

As per Kirchoff's Law,

$I_{}$ = $I_{1}$ + $I_{2}$ + $I_{3}$ ( ∵ $I_{}$ =  $\frac{V}{R_{eff} }$ )

$\frac{V}{R_{eff} }$  = V/R₁ + V/R₂ +  V/R₃

∴ $\frac{1}{R_eff} }$ = $\frac{1}{R_{1} }$ + $\frac{1}{R_{2} }$ + $\frac{1}{R_{3} }$  

   $R_{eff}$  = R₁R₂R₃/ (R₁R₂ + R₂R₃ + R₁R₃)