#### Answer

a. 2 A.
b. 1 A.
c. 3 A.
d. $4 \Omega$.

#### Work Step by Step

a. $I = \frac{V}{R} = \frac{12V}{6 \Omega} = 2 A$.
b. A pair of resistors wired in series has an equivalent resistance equal to the sum of the resistances. This is Rule 2 on page 443.
The two resistors, considered as a branch, then have an equivalent resistance of $12 \Omega$. $I = \frac{V}{R} = \frac{12V}{12 \Omega} = 1 A$.
c. The total current in the circuit, which is supplied by the voltage source, equals the sum of the currents in the branches (Rule 3, page 444). 2 A + 1 A = 3 A.
d. We effectively have a $12 \Omega$ resistor in parallel with a $6 \Omega$ resistor.
The equation is not given in this textbook, but for 2 resistors in parallel, the following relation holds.
$$\frac{1}{R_{parallel}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$$
Apply the equation to this problem.
$$\frac{1}{R_{parallel}} = \frac{1}{12 \Omega} + \frac{1}{6 \Omega } = \frac{3}{12 \Omega} $$
$$R_{parallel} = 4 \Omega$$
As a confirmation, one may also apply Ohm's Law and arrive at the same answer.
$$3 A = I = \frac{V}{R} = \frac{12V}{x}$$.
$$x = 4 \Omega$$.