# Three bulbs, each

a)1A, b)0.67A, c)0.33A

Given,

Emf of the battery, E=60V

Resistance of each bulb, r=180Ω

Formula used:

The potential difference across each bulb will be equal as they are connected in parallel across the same cell.

Also, we know that, for parallel connection of resistors, r with equal resistance, the effective resistance, Reff will be

Where n is the number of resistors.

a)

When all the switches are closed (switched on), the current will split equally across each of the resistors. And the total current will be the ratio between the potential difference and the effective resistance.

The effective resistance will be,

Or,

Hence the current,i will be

b)

In this case two resistors (or bulbs) are connected in parallel. So the effective resistance will be,

Or,

Hence the current, i will be

c)

In this case one resistor (or bulb) only connected to the battery. So the effective resistance will be,

Hence the current, i will be

The current in the setup due to the connection of 3 bulbs, 2 bulbs and 1 bulb is respectively 1.0A, 0.67A and 0.33A

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