When a certain a.c. supply is connected to a lamp, it lights with the same brightness as it does with a 12V battery

a) what is the r.m.s. value of the a.c. supply?
b) what is the peak p.d. of the a.c. supply?

Isn't rms value the same effective value as DC? What does root mean square give you if not that?

To find the r.m.s. (root mean square) value of an alternating current (a.c.) supply, we need to understand that the r.m.s. value represents the equivalent continuous current that would produce the same heating effect in a resistor as the a.c. supply.

a) To determine the r.m.s. value of the a.c. supply, we can equate it to the 12V battery, which has a constant and steady voltage output. This means that the r.m.s. value of the a.c. supply is also 12V.

b) The peak p.d. (peak voltage) of an a.c. supply can be determined using the formula:

Peak Voltage = R.m.s. value × √2

In this case, if the r.m.s. value of the a.c. supply is also 12V, we can substitute this value into the equation:

Peak Voltage = 12V × √2
Peak Voltage ≈ 16.97V (rounded to two decimal places)

So, the peak p.d. of the a.c. supply is approximately 16.97V.

Coming to your second question, the r.m.s. value does represent the effective value of an a.c. supply, just like it does for a DC (direct current) supply. The term "root mean square" characterizes the statistical average of a set of values. In the case of an a.c. supply, it calculates the square root of the mean value of the squared instantaneous values over one complete cycle.

This r.m.s. value is equivalent to the direct current that would produce the same average power dissipation in a resistor as the a.c. supply. Hence, you can consider the r.m.s. value as the effective value, which takes into account the changing magnitude and direction of the alternating current.