Question

Question 23:
In the circuit diagram given below five resistances of 10 Ω, 40 Ω, 30 Ω, 20 Ω and 60 Ω are connected as shown to a 12 V battery.

Calculate :
(a) total resistance in the circuit.
(b) total current flowing in the circuit.

Lakhmir Singh Physics Class 10

Answer 1

(a) step 1: 10Ω and 40Ω are joined in parallel combination.

so, R = 10 × 40/(10 + 40) = 8Ω

step 2 : 30Ω, 20Ω and 60Ω are joined in parallel combination.

so, 1/R' = 1/30 + 1/20 + 1/60 = 1/10

or, R' = 10Ω

step 3 : 8Ω and 10Ω are joined in series combination.

so, Req = R + R' = 8Ω + 10Ω = 18Ω

hence, equivalent resistance or total resistance = 18Ω

(b) total current flowing in the circuit , I = V/Req

given, emf of battery, V = 12 volts

so, I = 12/18 = 2/3 ≈ 0.67 Amp

Answer 2

Answer:

Total resistance or total resistance = 18Ω

Total current flowing in the circuit , I = $\frac{V}{Req}$ = 0.67 Amp

Explanation:

According to this question

step 1: 10Ω and 40Ω are joined in parallel combination.

So, R = $10\times \frac{40}{10 + 40}$ = 8Ω

Step 2: 30Ω, 20Ω and 60Ω are joined in parallel combination.

So, $\frac{1}{R'}$ = $\frac{1}{30}\ +\ \frac{1}{20}\ +\ \frac{1}{60}[/tex ]= [tex] \frac{1}{10}$

or, R' = 10Ω

Step 3 : 8Ω and 10Ω are joined in series combination.

So, Req = R + R' = 8Ω + 10Ω = 18Ω

Hence, equivalent resistance or total resistance = 18Ω

Total current flowing in the circuit , I = $\frac{V}{Req}$

given, emf of battery, V = 12 volts }

So, I = $\frac{12}{18}$ = $\frac{2}{3}$ ≈ 0.67 Amp