Question

name:________ period:________ date: 3/25/2023
series and parallel circuits practice with vir table
series circuits parallel circuits
$i = i_1=i_2 = i_3=ldots$ $i = i_1 + i_2+i_3+ldots$
$v = v_1 + v_2+v_3+ldots$ $v = v_1 = v_2 = v_3=ldots$
$r_{eq}=r_1 + r_2+r_3+ldots$ $\frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+ldots$
ohm’s law: v = i*r
1)
+9v
r1 2ω
r2 4ω

	v - voltage	i - current	r - resistance
r2	3	0.75	4
total	9	1.5	6

show work:
$q = i(t)$
$\frac{q}{t}=\frac{i(t)}{t}$
$1.5 = i$
$v=(0.75)(2)$
$v = 1.5$
2)
+6v
r1 1ω
r2 2ω
r3 3ω

	v - voltage	i - current	r - resistance
r2
r3
total

show work:

Explanation:

Step1: Calculate total resistance in series circuit

For the second - circuit with $R_1 = 1\Omega$, $R_2=2\Omega$ and $R_3 = 3\Omega$ in series, use the formula $R_{eq}=R_1 + R_2+R_3$. So $R_{eq}=1 + 2+3=6\Omega$.

Step2: Calculate total current in series circuit

Given $V = 6V$ and using Ohm's law $I=\frac{V}{R}$, with $R = R_{eq}=6\Omega$, then $I=\frac{6}{6}=1A$. In a series circuit, the current is the same throughout, so $I_1=I_2=I_3 = I = 1A$.

Step3: Calculate voltage across each resistor

For $R_1$: Using $V = IR$, with $I = 1A$ and $R_1=1\Omega$, then $V_1=1\times1 = 1V$.
For $R_2$: Using $V = IR$, with $I = 1A$ and $R_2=2\Omega$, then $V_2=1\times2 = 2V$.
For $R_3$: Using $V = IR$, with $I = 1A$ and $R_3=3\Omega$, then $V_3=1\times3 = 3V$.

	V - Voltage	I - Current	R - Resistance
R2	2V	1A	2Ω
R3	3V	1A	3Ω
total	6V	1A	6Ω

Answer:

	V - Voltage	I - Current	R - Resistance
R2	2V	1A	2Ω
R3	3V	1A	3Ω
total	6V	1A	6Ω