= **120** **V** **and** I = RMS current
I = P/VcosÏ† = **375**/(**120** x **0.72**) = 4.34 A
iv) Real **power** **is** also **given** by: P = I2R' where R' **is** **the** total resistance **in** **the** **circuit**
**375** = 4.342R'
R' = 19.91 ohm
**In** **the** **circuit** R = 10 ohm, hence 9.91

See Eq 3
Both **the** current **and** **power** **in** **the** **circuit** varies with **the** angular frequency **w**. As **w** appears **in** **the** impedance, **for** maximum current amplitude **and** average **power** **the** impedance of **the** **circuit** must be a minimum.

116621 Lightbulbs **in** a series/parallel Lightbulbs **in** a series/parallel (see attachment **for** complete solution **and** diagram)
a) Three bulbs **in** series :
**Power** **dissipated** **in** a resistance **is** **given** by **the** following expression :
P = VI = I^2R = **V**^2

d2) R1 = 100 â„¦, R2 = 300 â„¦, R3 = 200 â„¦
**In** **the** **circuit** diagram below, **the** total current flowing through **the** **circuit** **is** **1** Amp, calculate **the** total **power** using P = I2*R total, **and** then verify **the** answer by calculating **power** **dissipated** **in** each of **the** resistors

**In** order to get **power** **factor** **1**, . **The** **power** **factor** can be increased to be **1** by connecting a capacitance parallel to **the** old as shown **in** **figure**.
Now, RCL series **circuit** are analyzed.

P = V2/R = **120*****120**/(9.70*103) = 1.48 **W**
7.
**In** **the** **circuit** shown **in** **the** **figure** , **the** voltage across **the** 2.00 resistor **is** 12.5 .
What **is** **the** emf of **the** battery?

If a light bulb has half **the** resistance of a 100-**W** lightbulb, what would be its wattage? Assume both bulbs are attached to **the** same **120**-**V** **circuit**.

leaving **the** appliance on **for** this time
5 Days **is** time T = 5 x 24 = **120** h
**Power** **is** 1800 **W** so **in** **1** hour **the** dryer uses
E = 1800 x **1** = 1.8 KWh
**In** 5 days it would use
E(5) = 1.8 x **120** = 216 KWh
As **1** KWh costs 9.4 c
216 KWh will cost 216

We know that **power** **dissipated** **in** a resistor **is** give by P = I x **V** , where I **and** **V** are current **and** voltage on **the** resistor respectively. But this **is** **for** constant I **and** **V**.

348312 Current flowing **in** **the** series **circuit**, **Power** **dissipated**, DC current through **the** handset Current flowing **in** **the** series **circuit**, **Power** **dissipated**, DC current through **the** handset Please see **the** attachment as well **for** clear symbols **and** drawing **in** (