You don't say whether you're looking for the peak value of voltage or current.
-- The peak value of the sine is ' 1 ', so the peak voltage is 17 volts.
-- You haven't mentioned whether the load is complex or all real,
so naturally I'll assume it to be all real. Then the peak current is 17 volts/68 ohms = 0.25 amp.
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Peak to Peak is the most positive peak to the negative peak value. Or find any peak value and multiply by 2.
From ohms law, I = V/R hence Voltage and Resistance can affect the value of current, both peak and average. Also with a rectifier circuit other factors can affect the peak current such as frequency and capacitance Craig - AUT
Unless otherwise stated, the value of an a.c. current or voltage is expressed in r.m.s. (root mean square) values which, for a sinusoidal waveform, is 0.707 times their peak value. The output of a voltage (or potential) transformer is no different, its measured voltage will be its r.m.s value which is lower than its peak value.
To find the root mean square (rms) value for a voltage given in peak-to-peak (Vpp), you need to divide the Vpp value by 2√2. In this case, the Vpp is 300mV, which is equivalent to 0.3V. Dividing 0.3V by 2√2 ≈ 2.828, the rms value is approximately 0.106 V.
rms value of ac power = dc power in reference to heat production in pure resistive load So ac power of some rms value will produce the same heat in resistive load as dc power will of same value