The current in a simple single resistor circuit is I$1=6\mathrm{.}75$ A$.$ A second resistor, with resistance R$2=56$ ohms is added in series with the first and the current drops to I$2=2\mathrm{.}4$ A$.$

Part a$)$ Enter an expression for the resistance, R$1,$ of the first ressitor.

Part b$)$ calculate the resitance in ohms.

Part c$)$ Enter an expression for the resistance R$2$ in terms of I$2,$ R$1,$ and the voltage of the source, V$.$