Question

17. For the circuit shown in the diagram below:
4V
3 Ω
6Ω
12Ω
3Ω​

Answer 1

(i) R=6+3=9Ω

V=4V

I=

R

V

=

9

4

=0.45A

(ii) 12 ohm and 3 ohm are in series

R=12+3=15Ω

I=

R

V

=

15

4

=0.27A

p.d across 12 ohms= IR=0.27×12=3.24V

hope it helps you

please mark me a brainlist please