Calculate P1, P2, P3 and enter the values in the table. Find all the unknown values of E, I, R and P for the Parallel circuit above.

The total voltage $E_T$ is given as $9V$.
In a parallel circuit, the voltage across each resistor is equal to the total voltage.

$$V_1 = V_2 = E_T = 9V$$

We use Ohm's Law, $I = \frac{V}{R}$.
Given $R_1 = 1k\Omega = 1000\Omega$ and $R_2 = 10k\Omega = 10000\Omega$.

$$I_1 = \frac{V_1}{R_1} = \frac{9V}{1000\Omega} = 0.009A = 9mA$$ $$I_2 = \frac{V_2}{R_2} = \frac{9V}{10000\Omega} = 0.0009A = 0.9mA$$

The total current in a parallel circuit is the sum of the individual branch currents.

$$I_T = I_1 + I_2 = 0.009A + 0.0009A = 0.0099A = 9.9mA$$

We use the power formula, $P = V \cdot I$.

$$P_1 = V_1 \cdot I_1 = 9V \cdot 0.009A = 0.081W = 81mW$$ $$P_2 = V_2 \cdot I_2 = 9V \cdot 0.0009A = 0.0081W = 8.1mW$$

The total power is the sum of the power dissipated by each resistor.

$$P_T = P_1 + P_2 = 0.081W + 0.0081W = 0.0891W = 89.1mW$$

Alternatively, $P_T = E_T \cdot I_T = 9V \cdot 0.0099A = 0.0891W$.

For parallel resistors, the reciprocal of the total resistance is the sum of the reciprocals of individual resistances.

$$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{1000\Omega} + \frac{1}{10000\Omega}$$ $$\frac{1}{R_T} = \frac{10}{10000\Omega} + \frac{1}{10000\Omega} = \frac{11}{10000\Omega}$$ $$R_T = \frac{10000\Omega}{11} \approx 909.09\Omega = 0.909k\Omega$$

The completed table is:

	VCE (V)	I (mA)	R (k$\Omega$)	P (mW)
$R_1$	9	9	1	81
$R_2$	9	0.9	10	8.1
Total	9	9.9	0.909	89.1

The unknown values are:

• $V_1 = \boxed{\text{9 V}}$
• $I_1 = \boxed{\text{9 mA}}$
• $P_1 = \boxed{\text{81 mW}}$
• $V_2 = \boxed{\text{9 V}}$
• $I_2 = \boxed{\text{0.9 mA}}$
• $P_2 = \boxed{\text{8.1 mW}}$
• $I_T = \boxed{\text{9.9 mA}}$
• $R_T = \boxed{\text{0.909 k}\Omega}$
• $P_T = \boxed{\text{89.1 mW}}$