| Concept | Formula | Description |
|---|---|---|
| Ohm’s Law | $V = IR $ | Voltage, current, resistance relationship |
| Resistance (series) | $R_{total} = R_1 + R_2 + \dots $ | Total resistance in series |
| Resistance (parallel) | $\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots $ | Total resistance in parallel |
| Power | $P = IV $ | Power as product of current and voltage |
| Power from resistance | $P = I^2R $ | Power dissipated by a resistor |
| Power from voltage | $P = \frac{V^2}{R} $ | Alternative power formula using voltage |
The electron and proton in a hydrogen atom are separated by approximately $5.3 \times 10^{-11} $ m. If we could create a minuscule resistor between these particles with a resistance of $1.0 \times 10^6~\Omega $, what would be the current flow if the potential difference is $3.0 \times 10^{-19} $ V?
Parent formula:
$I = \frac{V}{R} $ (Ohm’s Law)
$V = 3.00 \times 10^{-19}~\text{V} $
$R = 1.00 \times 10^{6}~\Omega $
$I = \frac{V}{R} $
$I = \frac{3.00 \times 10^{-19}~\text{V}}{1.00 \times 10^{6}~\Omega} $
$I = 3.00 \times 10^{-25}~\text{A} $
Sig Figs: 3
A circuit contains a resistor of $5.00~\Omega $. When connected to a battery, it draws a current of $2.50~\text{A} $.
a) What is the voltage of the battery?
Parent formula:
$V = IR $
$I = 2.50~\text{A} $
$R = 5.00~\Omega $
</div>
<div class="calculations">
$V = IR $
$V = (2.50~\text{A})(5.00~\Omega) $
$V = 12.5~\text{V} $
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<div class="sigfig">
Sig Figs: 3<br>
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b) If a second identical resistor is added in series, what will be the new current?
<div class="parent-formula">
<b>Parent formula:</b><br>
$V = IR $<br> $R_{\text{series}} = R_1 + R_2 $
</div>
<div class="givens">
$V = 12.5~\text{V} $
$R_1 = 5.00~\Omega $, $R_2 = 5.00~\Omega $
</div>
<div class="calculations">
$R_{\text{total}} = 5.00~\Omega + 5.00~\Omega = 10.0~\Omega $
$I = \frac{V}{R_{\text{total}}} $
$I = \frac{12.5~\text{V}}{10.0~\Omega} $
$I = 1.25~\text{A} $
</div>
<div class="sigfig">
$I = 1.25~\text{A} $ </div>
c) If instead the second resistor is added in parallel with the first, what will be the new current?
Parent formula:
$V = IR $
$\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} $
$V = 12.5~\text{V} $
$R_1 = 5.00~\Omega $, $R_2 = 5.00~\Omega $
$R_{\text{parallel}} = \frac{1}{\frac{1}{5.00} + \frac{1}{5.00}} = 2.50~\Omega $
$I = \frac{V}{R_{\text{parallel}}} $
$I = \frac{12.5~\text{V}}{2.50~\Omega} $
$I = 5.00~\text{A} $
$I = 5.00~\text{A} $
Three resistors with values $6.00~\Omega $, $12.0~\Omega $, and $18.0~\Omega $ are connected in series with a $36.0~\text{V} $ battery. (a) Calculate the total resistance of the circuit.
Parent formula:
$R_{total} = R_1 + R_2 + R_3 $
$R_1 = 6.00~\Omega $
$R_2 = 12.0~\Omega $
$R_3 = 18.0~\Omega $
(a) $R_{total} = 6.00~\Omega + 12.0~\Omega + 18.0~\Omega = 36.0~\Omega $
$R_{total} = 36.0~\Omega $
(b) What is the current in the circuit? (c) What is the voltage across each resistor?
Parent formula:
$I = \frac{V}{R_{total}} $
$V_{battery} = 36.0~\text{V} $
$R_{total} = 36.0~\Omega $
(b) $I = \frac{36.0~\text{V}}{36.0~\Omega} = 1.00~\text{A} $
$I = 1.00~\text{A} $
Parent formula:
$V_n = IR_n $
$I = 1.00~\text{A} $
$R_1 = 6.00~\Omega $
$R_2 = 12.0~\Omega $
$R_3 = 18.0~\Omega $
$V_1 = 1.00~\text{A} \times 6.00~\Omega = 6.00~\text{V} $
$V_2 = 1.00~\text{A} \times 12.0~\Omega = 12.0~\text{V} $
$V_3 = 1.00~\text{A} \times 18.0~\Omega = 18.0~\text{V} $
$V_1 = 6.00~\text{V} $
$V_2 = 12.0~\text{V} $
$V_3 = 18.0~\text{V} $
In a string of 50 identical Christmas lights connected in series, each bulb has a resistance of $24~\Omega $. If the string is connected to a $120~\text{V} $ outlet: (a) What is the total resistance of the circuit? (b) What is the current through the circuit? (c) What is the voltage across each bulb? (d) What happens to the brightness of the remaining bulbs if one bulb burns out? Explain.
Parent formula:
$R_{total} = 50 \times R_{bulb} $
$I = \frac{V}{R_{total}} $
$V_{bulb} = \frac{V_{total}}{50} $
$R_{bulb} = 24~\Omega $
$V_{total} = 120~\text{V} $
$n = 50~\text{bulbs} $
(a) $R_{total} = 50 \times 24~\Omega = 1200~\Omega $
(b) $I = \frac{V}{R_{total}} = \frac{120~\text{V}}{1200~\Omega} = 0.100~\text{A} $
(c) $V_{bulb} = \frac{V_{total}}{50} = \frac{120~\text{V}}{50} = 2.40~\text{V} $
(d) If one bulb burns out, the circuit is broken and no current flows. All bulbs go out because the circuit has been opened. This is a disadvantage of series circuits for lighting.
$R_{total} = 1200~\Omega $
$I = 0.100~\text{A} $
$V_{bulb} = 2.40~\text{V} $
Three resistors with values $15.0~\Omega $, $30.0~\Omega $, and $45.0~\Omega $ are connected in parallel with a $12.0~\text{V} $ battery. (a) Calculate the total resistance of the circuit. (b) What is the current drawn from the battery? (c) What is the current through each resistor?
Parent formula:
$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $
$I_{total} = \frac{V}{R_{total}} $
$I_n = \frac{V}{R_n} $
$R_1 = 15.0~\Omega $
$R_2 = 30.0~\Omega $
$R_3 = 45.0~\Omega $
$V = 12.0~\text{V} $
(a) $\frac{1}{R_{total}} = \frac{1}{15.0~\Omega} + \frac{1}{30.0~\Omega} + \frac{1}{45.0~\Omega} $
$\frac{1}{R_{total}} = \frac{3.00}{45.0~\Omega} + \frac{1.50}{45.0~\Omega} + \frac{1.00}{45.0~\Omega} = \frac{5.50}{45.0~\Omega} $
$R_{total} = \frac{45.0~\Omega}{5.50} = 8.18~\Omega $
(b) $I_{total} = \frac{V}{R_{total}} = \frac{12.0~\text{V}}{8.18~\Omega} = 1.47~\text{A} $
(c) $I_1 = \frac{V}{R_1} = \frac{12.0~\text{V}}{15.0~\Omega} = 0.800~\text{A} $
$I_2 = \frac{V}{R_2} = \frac{12.0~\text{V}}{30.0~\Omega} = 0.400~\text{A} $
$I_3 = \frac{V}{R_3} = \frac{12.0~\text{V}}{45.0~\Omega} = 0.267~\text{A} $
Verification: $I_{total} = I_1 + I_2 + I_3$
$ 0.800~\text{A} + 0.400~\text{A} + 0.267~\text{A} = 1.47~\text{A} $
Sig Figs: 3
(Limited by voltage 12.0 V)
In a modern home, many outlets are connected in parallel to a $120~\text{V} $ source. If a circuit has a circuit breaker rated at $20~\text{A} $, calculate:
a) The minimum total resistance the circuit can have before the breaker trips.
Parent formula:
$I_{max} = 20~\text{A} $
$R_{min} = \frac{V}{I_{max}} $
$P = IV $
$P_{available} = P_{max} - P_{used} $
$V = 120~\text{V} $
$I_{max} = 20~\text{A} $
$P_{heater} = 1500~\text{W} $
(a) $R_{min} = \frac{V}{I_{max}} = \frac{120~\text{V}}{20~\text{A}} = 6.0~\Omega $
(b) Maximum power the circuit can handle:
$P_{max} = V \times I_{max} = 120~\text{V} \times 20~\text{A} =$
$ 2400~\text{W} $
Available power for additional appliance:
$P_{available} = P_{max} - P_{heater} $
$2400~\text{W} - 1500~\text{W} = 900~\text{W} $
$R_{min} = 6.0~\Omega $
$P_{available} = 900~\text{W} $
For each of the following statements, identify whether it applies to series circuits, parallel circuits, or both:
a) The current through each component is the same.
Answer: Series circuits - In a series circuit, the current must be the same through each component because there is only one path for charge to flow.
b) The voltage across each component is the same.
Answer: Parallel circuits - Components connected in parallel all have the same voltage across them because they are connected directly to the same two points in the circuit.
c) If one component fails and becomes an open circuit, all other components stop working.
Answer: Series circuits - In a series circuit, if one component fails, the circuit is broken and current stops flowing through all components.
d) Adding more components decreases the total resistance.
Answer: Parallel circuits - In a parallel circuit, adding another resistance in parallel provides another path for current, reducing the total resistance.
e) The total resistance is always greater than the largest individual resistance.
Answer: Series circuits - In a series circuit, the total resistance is the sum of all individual resistances, so it must be greater than the largest individual resistance.
Consider two lightbulbs, one rated at 60 W and another at 100 W, both designed for use with 120 V.
a) Which bulb has higher resistance?
Parent formula:
$R = \frac{V^2}{P} $
$P = I^2R $
$P_1 = 60~\text{W} $
$P_2 = 100~\text{W} $
$V = 120~\text{V} $
$R_{60W} = \frac{V^2}{P} = \frac{(120~\text{V})^2}{60~\text{W}} = 240~\Omega $
$R_{100W} = \frac{V^2}{P} = \frac{(120~\text{V})^2}{100~\text{W}} = 144~\Omega $
The 60W bulb has higher resistance.
Sig Figs: 3
(Limited by voltage 120 V)
b) If these bulbs are connected in series to a 120 V source, which bulb will be brighter? Explain.
In series, both bulbs will have the same current. Since $P = I^2R $ and the 60W bulb has higher resistance, it will dissipate more power and be brighter than the 100W bulb (contrary to their ratings at 120V).
c) If these bulbs are connected in parallel to a 120 V source, which bulb will be brighter? Explain.
In parallel, both bulbs will receive the full 120V. The 100W bulb will draw more current and operate at its rated 100W, making it brighter than the 60W bulb.