Question

r1 = 3 kω
9 v
r2 = 10 kω
r3 = 5 kω
v(v) i(a) r(ω) p(w)
1
2
3
t
r3 3 ω

Explanation:

Step1: Calculate total resistance

In a series - circuit, the total resistance $R_T$ is the sum of individual resistances. $R_1 = 3\times10^{3}\Omega$, $R_2 = 10\times10^{3}\Omega$, $R_3 = 5\times10^{3}\Omega$. So $R_T=R_1 + R_2+R_3=(3 + 10+5)\times10^{3}\Omega=18\times10^{3}\Omega$.

Step2: Calculate the total current

Using Ohm's law $I=\frac{V}{R}$, where $V = 9V$ and $R = R_T=18\times10^{3}\Omega$. So $I_T=\frac{9V}{18\times10^{3}\Omega}=0.5\times10^{- 3}A = 0.5mA$.
In a series - circuit, the current through each resistor is the same, so $I_1 = I_2 = I_3=I_T = 0.5\times10^{-3}A$.

Step3: Calculate the voltage across each resistor

Using Ohm's law $V = IR$.
For $R_1$: $V_1=I_1R_1=(0.5\times10^{-3}A)\times(3\times10^{3}\Omega)=1.5V$.
For $R_2$: $V_2=I_2R_2=(0.5\times10^{-3}A)\times(10\times10^{3}\Omega)=5V$.
For $R_3$: $V_3=I_3R_3=(0.5\times10^{-3}A)\times(5\times10^{3}\Omega)=2.5V$.

Step4: Calculate the power dissipated in each resistor

Using the formula $P = VI$.
For $R_1$: $P_1=V_1I_1=(1.5V)\times(0.5\times10^{-3}A)=0.75\times10^{-3}W = 0.75mW$.
For $R_2$: $P_2=V_2I_2=(5V)\times(0.5\times10^{-3}A)=2.5\times10^{-3}W = 2.5mW$.
For $R_3$: $P_3=V_3I_3=(2.5V)\times(0.5\times10^{-3}A)=1.25\times10^{-3}W = 1.25mW$.
The total power $P_T=V_TI_T=(9V)\times(0.5\times10^{-3}A)=4.5\times10^{-3}W = 4.5mW$, and also $P_T=P_1 + P_2+P_3=(0.75 + 2.5+1.25)\times10^{-3}W=4.5mW$.

Answer:

	V(V)	I(A)	R(Ω)	P(W)
2	5	$0.5\times10^{-3}$	$10\times10^{3}$	$2.5\times10^{-3}$
3	2.5	$0.5\times10^{-3}$	$5\times10^{3}$	$1.25\times10^{-3}$
T	9	$0.5\times10^{-3}$	$18\times10^{3}$	$4.5\times10^{-3}$