Charge cannot accumulate at a junction; therefore the algebraic sum of currents at a junction is zero.
\[
\sum I{\text{in}} = \sum I{\text{out}} \qquad\text{or}\qquad I{\text{total}} = I{1}+I{2}+I{3}+\dots
\]
When components are connected end‑to‑end the same current flows through each. The total supplied p.d. is shared between the components:
\[
V{\text{total}} = V{1}+V{2}+V{3}+\dots
\]
These shortcuts are useful when you need the voltage across, or the current through, a particular component without solving the whole circuit.
| Situation | Formula | When to Use |
|---|---|---|
| Voltage across resistor \(R_{k}\) in a series string | \(V{k}=V{\text{total}}\displaystyle\frac{R_{k}}{\sum R}\) | Series circuit where the total p.d. is known. |
| Current through resistor \(R_{k}\) in a parallel network | \(I{k}=I{\text{total}}\displaystyle\frac{R{\text{eq}}}{R{k}}\) | Parallel circuit where the total current is known. |
| Arrangement | Formula |
|---|---|
| Series | \(R{\text{eq}} = R{1}+R{2}+R{3}+\dots\) |
| Parallel | \(\displaystyle\frac{1}{R{\text{eq}}}= \frac{1}{R{1}}+\frac{1}{R{2}}+\frac{1}{R{3}}+\dots\) |
| Quantity | Symbol | Formula | Typical Use |
|---|---|---|---|
| Current (junction rule) | I | \(I{\text{total}} = \sum I{\text{branch}}\) | Parallel circuits |
| Potential difference (series rule) | V | \(V{\text{total}} = \sum V{\text{component}}\) | Series circuits |
| Potential difference (parallel rule) | V | \(V{\text{parallel}} = V{\text{any branch}}\) | Parallel circuits |
| Resistance (series) | Req | \(R{\text{eq}} = R{1}+R_{2}+…\) | Total resistance in series |
| Resistance (parallel) | Req | \(\displaystyle\frac{1}{R{\text{eq}}}= \frac{1}{R{1}}+\frac{1}{R_{2}}+…\) | Total resistance in parallel |
| Ohm’s law | V, I, R | \(V = I R\) | Relate any two of V, I, R |
| Power | P | \(P = I V = I^{2}R = \dfrac{V^{2}}{R}\) | Calculate energy loss or heating |
| Voltage‑division (series) | Vk | \(V{k}=V{\text{total}}\dfrac{R_{k}}{\sum R}\) | Find voltage across a single series resistor |
| Current‑division (parallel) | Ik | \(I{k}=I{\text{total}}\dfrac{R{\text{eq}}}{R{k}}\) | Find current through a single parallel resistor |
Problem: A 12 V battery supplies a circuit consisting of a 4 Ω resistor \(R{1}\) in series with a parallel combination of \(R{2}=6\ \Omega\) and \(R_{3}=12\ \Omega\). Find:
Solution
Parallel equivalent:
\[
\frac{1}{R_{23}} = \frac{1}{6} + \frac{1}{12}= \frac{2}{12}+\frac{1}{12}= \frac{3}{12}
\qquad\Rightarrow\qquad
R_{23}= \frac{12}{3}=4\ \Omega
\]
Total resistance (series):
\[
R{\text{total}} = R{1}+R_{23}=4\ \Omega+4\ \Omega = 8\ \Omega
\]
Battery current (Ohm’s law):
\[
I{\text{total}} = \frac{V}{R{\text{total}}}= \frac{12\ \text{V}}{8\ \Omega}=1.5\ \text{A}
\]
Voltage across the parallel section (series rule):
\[
V{23}= I{\text{total}}\times R_{23}=1.5\ \text{A}\times4\ \Omega = 6\ \text{V}
\]
Currents in the parallel branches (junction rule or current‑division):
\[
I{2}= \frac{V{23}}{R_{2}}= \frac{6\ \text{V}}{6\ \Omega}=1.0\ \text{A}
\qquad
I{3}= \frac{V{23}}{R_{3}}= \frac{6\ \text{V}}{12\ \Omega}=0.5\ \text{A}
\]
Check: \(I{2}+I{3}=1.0+0.5=1.5\ \text{A}=I_{\text{total}}\) – junction rule satisfied.
Voltage across each resistor (Ohm’s law):
All three branches have the same p.d., confirming the parallel‑voltage rule.
Power dissipated:
\[
P{R1}=I{\text{total}}^{2}R_{1}= (1.5)^{2}\times4 = 9.0\ \text{W}
\]
\[
P{R2}=I{2}^{2}R_{2}= (1.0)^{2}\times6 = 6.0\ \text{W}
\]
\[
P{R3}=I{3}^{2}R_{3}= (0.5)^{2}\times12 = 3.0\ \text{W}
\]
Total power = 18 W, which equals \(V{\text{battery}}I{\text{total}} = 12\ \text{V}\times1.5\ \text{A}\).
Parallel equivalent:
\[
\frac{1}{R{\text{eq}}}= \frac{1}{3}+ \frac{1}{6}= \frac{2}{6}+ \frac{1}{6}= \frac{3}{6}\;\Rightarrow\;R{\text{eq}}=2\ \Omega
\]
Total current:
\[
I_{\text{total}}= \frac{9\ \text{V}}{2\ \Omega}=4.5\ \text{A}
\]
Branch currents (Ohm’s law or current‑division):
\[
I_{1}= \frac{9\ \text{V}}{3\ \Omega}=3.0\ \text{A},\qquad
I_{2}= \frac{9\ \text{V}}{6\ \Omega}=1.5\ \text{A}
\]
Power:
\[
P{1}=I{1}^{2}R_{1}=3.0^{2}\times3=27\ \text{W},\;
P{2}=I{2}^{2}R_{2}=1.5^{2}\times6=13.5\ \text{W}
\]
Total resistance (series):
\[
R_{\text{total}}=2+4+8=14\ \Omega
\]
Current:
\[
I=\frac{12\ \text{V}}{14\ \Omega}=0.86\ \text{A}\;(2\ \text{sf})
\]
Voltages (voltage‑division):
\[
V{1}=I R{1}=0.86\times2=1.7\ \text{V}
\]
\[
V{2}=I R{2}=0.86\times4=3.4\ \text{V},\qquad
V{3}=I R{3}=0.86\times8=6.9\ \text{V}
\]
Power (optional):
\[
P{1}=I^{2}R{1}=0.86^{2}\times2=1.48\ \text{W}
\]
\[
P{2}=I^{2}R{2}=0.86^{2}\times4=2.96\ \text{W}
\]
\[
P{3}=I^{2}R{3}=0.86^{2}\times8=5.92\ \text{W}
\]
Parallel equivalent of 10 Ω and 15 Ω:
\[
\frac{1}{R_{p}}= \frac{1}{10}+ \frac{1}{15}= \frac{3}{30}+ \frac{2}{30}= \frac{5}{30}
\qquad\Rightarrow\qquad
R_{p}=6\ \Omega
\]
Total resistance (series):
\[
R_{\text{total}} = 5\ \Omega + 6\ \Omega = 11\ \Omega
\]
Total current from the battery:
\[
I_{\text{total}}= \frac{24\ \text{V}}{11\ \Omega}=2.18\ \text{A}
\]
Voltage across the parallel section (series rule):
\[
V{p}= I{\text{total}}\times 6\ \Omega = 13.1\ \text{V}
\]
Current through the 10 Ω resistor (current‑division):
\[
I{10}= I{\text{total}}\frac{R_{p}}{10}=2.18\frac{6}{10}=1.31\ \text{A}
\]
Voltage across the 5 Ω resistor:
\[
V{5}= I{\text{total}}\times5\ \Omega = 10.9\ \text{V}
\]
If the 15 Ω resistor is removed, the parallel part becomes a single 10 Ω resistor, so \(R_{\text{eq}}=10\ \Omega\). New total resistance \(=5+10=15\ \Omega\) and the total current falls to \(\displaystyle I'=\frac{24}{15}=1.60\ \text{A}\). Thus the current decreases.
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