Question

A voltmeter reading of a circuit it was 6v. The resistance of the bulb was 3 ohms

I. What will be the reading of the ammeter
II. 5 ohm resistor is then added in series what is the reading of the ammeter now
III. If the 5 ohm resistor was added in parallel to the ammeter.

2. Three resistance of 1ohm,3ohm and 4ohm respectively are connected in series to a 12v battery of negligible internal resistance
Calculate the current flowing through each resistor​

Answer 1

Step-by-step explanation:

I. Ohm's law:

V = IR

6 v = I (3 Ω)

I = 2 A

II. Resistors in series:

R = R₁ + R₂

R = 3 Ω + 5 Ω

R = 8 Ω

V = IR

6 v = I (8 Ω)

I = 0.75 A

III. Resistors in parallel:

1/R = 1/R₁ + 1/R₂

1/R = 1 / (3 Ω) + 1 / (5 Ω)

R = 1.875 Ω

V = IR

6 v = I (1.875 Ω)

I = 3.2 A

2. Resistors in series:

R = R₁ + R₂ + R₃

R = 1 Ω + 3 Ω + 4 Ω

R = 8 Ω

V = IR

12 v = I (8 Ω)

I = 1.5 A