A

Step 1: Given the e.m.f. expression:
$$ E(t) = 150 \, \cos(314t) \text{ volts} $$

Step 2: We know that the current through a purely resistive circuit is given by Ohm's law:
$$ I(t) = \frac{E(t)}{R} $$
Substituting the values we have:
$$ I(t) = \frac{150 \, \cos(314t)}{30} $$
Simplifying this gives:
$$ I(t) = 5 \, \cos(314t) \text{ amperes} $$

Step 3: Now, we need to find the power as a function of time. The instantaneous power consumed in a resistive circuit can be expressed as:
$$ P(t) = I(t)^2 \cdot R $$
Substituting the expression of current we found earlier:
$$ P(t) = (5 \, \cos(314t))^2 \cdot 30 $$
This simplifies to:
$$ P(t) = 25 \, \cos^2(314t) \cdot 30 $$
$$ P(t) = 750 \, \cos^2(314t) \text{ watts} $$
We can now use the trigonometric identity for cosine squared: $$ \cos^2(x) = \frac{1 + \cos(2x)}{2} $$
So we replace $$ \cos^2(314t) $$:
$$ P(t) = 750 \cdot \frac{1 + \cos(628t)}{2} $$
$$ P(t) = 375 + 375 \cdot \cos(628t) \text{ watts} $$

Step 4: Now, we can determine the frequencies of current and power variations. The frequency can be found from the argument of the cosine function which is of the form $$ \cos(\omega t) $$:
- For current, $$ \omega_{current} = 314 \implies f_{current} = \frac{314}{2\pi} \approx 50 \text{ Hz} $$
- For power, $$ \omega_{power} = 628 \implies f_{power} = \frac{628}{2\pi} \approx 100 \text{ Hz} $$

Final Answers:
The expressions for current and power as a function of time are:
$$ I(t) = 5 \, \cos(314t) \text{ amperes} $$
$$ P(t) = 375 + 375 \cdot \cos(628t) \text{ watts} $$
The frequency of current variations is approximately 50 Hz and the frequency of power variations is approximately 100 Hz.