The Current I Through a Resistor of Pure Resistance 2

The current i through a resistor of pure resistance 2 kΩ varies with time t according to the equation i = 5 sin (2π x 100t + 45º) mA. The voltage across the resistor will thus be:

A) v = 5 sin (2 π % 100t + 45º) V
B) v = 5 sin (2 π % 100t - 45º) V
C) v = 10 sin (2 π % 100t + 45º) V
D) v = 10 sin (2 π % 100t + 135º) V