Question

Resistance

Current

Voltage

𝑅1

3kΩ

𝟓 𝒙 𝟏𝟎−𝟒𝑨

1.5 V

𝑅2

10 kΩ

𝟓 𝒙 𝟏𝟎−𝟒𝑨

5V

𝑅3

5kΩ

𝟓 𝒙 𝟏𝟎−𝟒𝑨

2.5 V

Total

18 kΩ

𝟓 𝒙 𝟏𝟎−𝟒𝑨

Answer 1

Given,

$R_1=3k\Omega$

$I_1=5\times 10^{-4}A$

$V_1=1.5V$

$R_2=10k\Omega$

$I_2=5\times 10^{-4}A$

$V_2=5V$

$R_3=5k\Omega$

$I_3=5\times 10^{-4}A$

$V_3=2.5V$

As here we can see that the current in all the three resistance is same and the potential at all the three resistance is different, hence we can conclude that all the three resistances are connected in series.

Hence, equivalent resistance $(R_{eq})=R_1+R_2+R_3$

$=3 k\Omega+ 10k\Omega+ 5k\Omega$

$=18k\Omega$

Net applied potential $N_{net}=V_1+V_2+V_3$

$=1.5V+5V+2.5V$

$=9V$