Given that,
Three resistors of resistances 2 Ω, 3 Ω and 4 Ω are connected in series with a 15 V supply.
Solution,
The equivalent resistance in series combination is given by :
[tex]R=R_1+R_2+R_3\\\\R=2+3+4\\\\=9\ \Omega[/tex]
(i) Total resistance of the circuit is equal to 9 ohms.
(ii) Let I be the total current. Using Ohm's law as follows :
[tex]V=IR\\\\I=\dfrac{V}{R}\\\\I=\dfrac{15}{9}\\\\=1.67\ A[/tex]
(iii) Voltage drop across 2 ohm resistor is :
[tex]V=1.67\times 2=3.34\ V[/tex]
Voltage drop across 3 ohm resistor is :
[tex]V=1.67\times 3=5.01\ V[/tex]
Voltage drop across 4 ohm resistor is :
[tex]V=1.67\times 4=6.68\ V[/tex]
Hence, this is the required solution.
