Question

1. Four identical lamps are connected in parallel to a 6 volt battery. What is the voltage drop across each lamp?
2. What happens to the total resistance in a network when more resistors are added to a series circuit ?
3. Four resistors of resistances 3Ω , 4Ω , 5Ω and 6Ω are connected in parallel to a 12 volts source, what is the current through the 5Ω resistor.

Answer 1

Gives four identical lamp resistance $R=R_1=R_2=R_3=R_3=R_4$

All resistance is parallel combination

$R_{net}=\frac{R}{4}$

Voltage drop across each resistance are equal

$V=$ 6V

$=V=V_1=V_2=V_3=V_4=6V$

(2)

Four resistance is series combination

$R_{net}=R_1+R_2+R_3+R_4$

R_net =3+4+5+6=14 $\Omega$

(3)

All four resistance are parallel combination

$\frac{1}{R}=\frac{20+15+12+10}{60}=\frac{57}{60}$

$R=\frac{60}{57}=1.05\Omega$

V=12V

$I=\frac{12}{1.05}=11.4A$

R=5ohm ResistanceCurrent through the find out

Ratio of current

$$I_1:I_2:I_3:I_4=\frac{1}{R_1}:\frac{1}{R_2}:\frac{1}{R_3}:\frac{1}{R_4}:$$

Put Value of all resistance

Ratio of currents

$I_1:I_2:I_3:I_4=20:15:12:10$

$I_3=\frac{12}{57}\times11.4$ =2.4A