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.​

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,

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,

Current flowing in 2.3 ohm resistor,

In parallel combination, are brighter than bulbs in series.

what are the statements? x

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.