Question 11.SP.15: A 20-Ω electric iron and a 100-Ω lamp are connected in paral......
A 20-Ω electric iron and a 100-Ω lamp are connected in parallel across a 120-V 60-Hz ac line (Fig. 11-26). Find the total current, the total resistance, and the total power drawn by the circuit, and draw the phasor diagram.
For a parallel circuit, V_{T} = V_{1} = V_{2} = 120\ V.
I_{1}=\frac{V_{1}}{R_{1}}=\frac{120}{20}=6\ A\ \ \ \ \ \ \ \ I_{2}=\frac{V_{2}}{R_{2}}=\frac{120}{100}=1.2\ AThen I_{T} = I_{1} + I_{2} = 6+1.2=7.2\ A
R_{T} = \frac{V_{T}}{I_{T}}=\frac{120}{7.2}=16.7\ \OmegaIn a purely resistive set of branch currents, the total current I_{T} is in phase with the total voltage V_{T}. The phase angle is therefore equal to 0°.
P = V_{T}I_{T} \cos \theta = 120(7.2)(\cos 0^{\circ}) = 120(7.2)(1) = 864\ WSince the voltage in a parallel circuit is constant, the voltage is used as the reference phasor. The currents I_{1} and I_{2} are drawn in the same direction as the voltage because the current through pure resistances is in phase with the voltage. The I_{1} phasor is shown longer than I_{2} because its current value is higher (see the phasor diagram).
