Question

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.

Answer 1

Explanation:

If all the resisters are connected in series

then Re=R1+R2+R3+R4+R5

Re=10+40+30+20+60

Re=160ohm

ReI=V

12=160/I

I=160/12

I= 40/3

therefore I is 13.33ampere

If all the resistors are connected in parallel then

1/Re=1/R1+1/R2+1/R3+1/R4+1/R5

Answer 2

Total Resistance in Circuit is 18 $\Omega$

Explanation:

Given,

$R_1 = 10 \Omega$

$R_2 = 40\ \Omega$

$R_3 = 30\ \Omega$

$R_4 = 20\ \Omega$

$R_5 = 60\ \Omega$

Voltage = 12 V

*Refer attached figure

a.) Total Resistance

 (Parallel Connection)

= $\frac {1}{10} + \frac {1}{40}$

$R_ P = 8\ \Omega$

(Parallel Connection)

= $\frac {1}{30} + \frac {1}{20} + \frac {1}{60}$

$R _P = 10 \Omega$

Total resistance = (Series Connection)

b.) V = IR (Ohm's Law)

$12 = \frac {I}{18}$

or $I = \frac {12}{18}$

I = 0.67 A