Question

Observe the diagram and answer following questions :-

i) Determine the ammeter reading. (2)

ii) Power consumed by 40 Ω resistor. (2)

iii) Potential difference across 20 Ω resister​

Answer 1

Question

Observe the diagram and answer the following questions :-

i) Determine the ammeter reading

ii)Power consumed by 40Ω resistor

iii) Potential difference across 20Ω resistor

ANSWER

Given : -

The EMF or applied voltage = 18 volts

Required to find : -

• i) Determine the ammeter reading
• ii)Power consumed by 40Ω resistor
• iii) Potential difference across 20Ω resistor

Formula used : -

Equivalent resistance when the resistors are connected in series

R_eq = R1+R2 ......

Equivalent resistance when the resistors are connected in parallel

[1]/[R_eq] = [1]/[R1]+[1]/[R2] ....

(or)

R_eq = (R1*R2)/(R1+R2+) ....

Solution : -

The Applied voltage in the given circuit = 18V

However,

From the figure (circuit) we can conclude that;

• 10Ω & 40Ω are connected in parallel to each other
• 30Ω , 20Ω & 60Ω resistors are connected in parallel to each other
• The equivalent resistance of the above 2 connections will be in series to each other .

Using this concept as the base let's crack the Question.

Since,

10Ω & 40Ω are connected in parallel to each other

This implies;

• R_eq = (R1*R2)/(R1+R2+) ....

R_eq = (10Ω*40Ω)/(10Ω+40Ω)

R_eq = (400Ω²)/(50Ω)

R_eq(1) = 8Ω

Similarly,

30Ω , 20Ω & 60Ω resistors are connected in parallel to each other

This implies;

R_eq = (30*20*60)/(30+20+60)

R_eq = (36000)/(110Ω)

R_eq = (3600)/(11)

R_eq = 327.2727....

R_eq(2) ≈ 327.3Ω

Now,

The equivalent resistance of the above 2 connections will be in series to each other .

R_eq(circuit) = 8Ω+327.3Ω

R_eq(circuit) = 335.3Ω

Using the Ohm's law

voltage (V) = current (I) x resistance (R)

18 = I x 335.3

I = (18)/(335.3)

I = 0.5368.....

I ≈ 0.54 amphere (A)

(i) Ammeter readings = 0.54 A

Now,

We know that !

In a parallel connection the current get's splited up.

So,

The resistors which are in parallel are 10Ω & 40Ω

This implies;

Current passing or flowing through 40Ω resistor is

V = 0.54 x 8Ω

V = 4.32 V

Potential difference across the ends is 4.32 volts

So,

I = ∆V/R

I = 18-4.32/40

I = 13.68/40

I = 0.342 A

Current flowing through 40Ω is 0.342 A

Power = current² (I²) x Resistance (R)

Thus,

Power = (0.342)² x 40

Power = 4.43232 A²/Ω

• Power consumed by 40Ω resistor ≈ 4.432 watt

From the above calculations it is clear that ,

The value of potential difference is

V = 0.54 x 20Ω

V = 10.8

The potential difference across 20Ω = 10.8 V