Question

Two resistors with resistance values of 4.5 Ω and 2.3 Ω are connected in series or parallel

across a 30 V potential difference to a light bulb.

a. Calculate the current delivered through the light bulb in the two cases.

b. Draw the circuit connection that will achieve the brightest light bulb.​

Answer 1

Step-by-step explanation:

Given that,

Two resistors 4.5 Ω and 2.3 Ω .

Potential difference = 30 V

When they are in series, the current through each resistor remains the same. First find the equivalent resistance.

R' = 4.5 + 2.3

= 6.8 Ω

Current,

`I=(V)/(R')\\\\I=(30)/(6.8)\\\\=4.41\ A`

So, the current through both lightbulb is the same i.e. 4.41 A.

When they are in parallel, the current divides.

Current flowing in 4.5 resistor,

`I_1=(V)/(R_1)\\\\=(30)/(4.5)\\\\I_1=6.7\ A`

Current flowing in 2.3 ohm resistor,

`I_2=(V)/(R_2)\\\\=(30)/(2.3)\\\\I_2=13.04`

In parallel combination, are brighter than bulbs in series.