Question

A series combination of 12 Ω and 3 Ω is connected in parallel with another series combination of 6 Ω and 3 Ω. If a potential difference of 4 V is applied across it find the
i) current drawn from the battery
ii) current through 12 Ω resistor
iii)potential difference across 6 Ω resistor?


Pls answer this question ASAP...... don’t insert fake ones.....thank you..........

Answer 1

Answer:

i) 1.16 A

ii) 1.16 A

iii) 9.96 V

Explanation:

R1 = 12 ohm

R2 = 3 ohm

R3 = 6 ohm

R4 = 3 ohm

V = 4 V

Rs1 = R1+R2

= 12+3

Rs1 = 15 ohm

Rs2 = R3+R4

= 6+3

Rs2 = 9 ohm

Rp = Rs1+Rs2

= 15+9

Rp = 24 ohm

i) We know that,

V = RI

I = V/R

= 4/24

= 1/6 A (or) 1.66

ii) 1.66 A because the value of "I" doesn't change in the parallel resistor

iii) V = RI

= 6 x 1.66

V = 9.96 V

Answer 2

Answer:

1.16 A

1.16A

9.9 v

dharm of a