Question

heya ...
solve all the points ..
it's urgent ......

Answer 1

Equivalent R between 6Ω and 3Ω resistor = R1R2/(R1+R2) = 18/9 = 2Ω
Total resistance of circuit = 1+4+1+2+2 = 10Ω===> R = 10Ω
(1) Eq. resistance will be 10Ω
V = 30V (given)
V = IR (ohm's law)==> I = V/R = 30/10 = 3amps

(2) The reading on the blue ammeter would be 3A.

(3) In 4Ω resistor, here R=4ΩV = IR = 3 * 4 = 12V
So the potential drop will be 30-12 = 18V.
In 6Ω resistor and 3Ω resistor there'll be same voltage drop since they are connected in parallel.
So R = eq.Resistance = 2Ω in this case
V = IR = 3 * 2 = 6V
So the potential drop will be 30-6 = 24 V

(4) In this case R = 6Ω and V = 6V (from (3))
V = IR
===> I = V/R = 6/6 = 1A

So the reading on the black ammeter will be 1A.

(5) Terminal voltage is 30V (from (1))
(6) There is no potential drop across the cell
(7) Reading of the voltmeter will be 12V (from (3))

Hope it helps!