QQuestionPhysics
QuestionPhysics
Determine the total resistance of each of the following parallel circuits. Then use Gizmo to check your answer. (You can calculate the total resistance from the current and voltage using Ohm’s law, or use the Ohmmeter to measure the resistance directly.)
A parallel circuit with a 20 -ohm resistor and a 10 -ohm resistor.
A parallel circuit with two 20 -ohm resistors and a 10 -ohm resistor.
A parallel circuit with a 15 -ohm light bulb and a 20 -ohm resistor.
A parallel circuit with two 100 -ohm resistors and a 20 -ohm resistor.
A parallel circuit with a 10 -ohm, 20 -ohm, 100 -ohm and 200 -ohm resistor
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Answer
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Step 1:: Recall the formula for calculating the total resistance in a parallel circuit.
where $R_1, R_2, R_3, \dots, R_n$ are the individual resistances in the circuit.
The total resistance in a parallel circuit is given by the reciprocal of the sum of the reciprocals of each individual resistance:
Step 2:: Calculate the total resistance for the first parallel circuit.
R_{total} = \frac{1}{\frac{1}{20 \ \Omega} + \frac{1}{10 \ \Omega}} = \frac{1}{\frac{1}{20} + \frac{1}{10}} \ \Omega
For the first parallel circuit with a 20 -ohm resistor and a 10 -ohm resistor, we have: Plugging these values into the formula, we get:
Step 3:: Simplify the expression.
\frac{1}{\frac{1}{20} + \frac{1}{10}} = \frac{1}{\frac{1}{20} + \frac{2}{20}} = \frac{1}{\frac{3}{20}} = \frac{20}{3} \ \Omega
Step 4:: Calculate the total resistance for the second parallel circuit.
R_{total} = \frac{1}{\frac{1}{20 \ \Omega} + \frac{1}{20 \ \Omega} + \frac{1}{10 \ \Omega}} = \frac{1}{\frac{1}{20} + \frac{1}{20} + \frac{1}{10}} \ \Omega
For the second parallel circuit with two 20 -ohm resistors and a 10 -ohm resistor, we have: Plugging these values into the formula, we get:
Step 5:: Simplify the expression.
\frac{1}{\frac{1}{20} + \frac{1}{20} + \frac{1}{10}} = \frac{1}{\frac{1}{20} + \frac{1}{10}} = \frac{1}{\frac{1}{20} + \frac{2}{20}} = \frac{1}{\frac{3}{20}} = \frac{20}{3} \ \Omega
Step 6:: Calculate the total resistance for the third parallel circuit.
R_{total} = \frac{1}{\frac{1}{15 \ \Omega} + \frac{1}{20 \ \Omega}} = \frac{1}{\frac{4}{60} + \frac{3}{60}} \ \Omega
For the third parallel circuit with a 15 -ohm light bulb and a 20 -ohm resistor, we have: Plugging these values into the formula, we get:
Step 7:: Simplify the expression.
\frac{1}{\frac{4}{60} + \frac{3}{60}} = \frac{1}{\frac{7}{60}} = \frac{60}{7} \ \Omega
Step 8:: Calculate the total resistance for the fourth parallel circuit.
R_{total} = \frac{1}{\frac{1}{100 \ \Omega} + \frac{1}{100 \ \Omega} + \frac{1}{20 \ \Omega}} = \frac{1}{\frac{1}{100} + \frac{1}{100} + \frac{1}{20}} \ \Omega
For the fourth parallel circuit with two 100 -ohm resistors and a 20 -ohm resistor, we have: Plugging these values into the formula, we get:
Step 9:: Simplify the expression.
\frac{1}{\frac{1}{100} + \frac{1}{100} + \frac{1}{20}} = \frac{1}{\frac{1}{100} + \frac{1}{50}} = \frac{1}{\frac{2}{100} + \frac{1}{50}} = \frac{1}{\frac{2}{100} + \frac{2}{100}} = \frac{1}{\frac{4}{100}} = \frac{100}{4} = 25 \ \Omega
Step 10:: Calculate the total resistance for the fifth parallel circuit.
R_{total} = \frac{1}{\frac{1}{10 \ \Omega} + \frac{1}{20 \ \Omega} + \frac{1}{100 \ \Omega} + \frac{1}{200 \ \Omega}}
For the fifth parallel circuit with a 10 -ohm, 20 -ohm, 100 -ohm and 200 -ohm resistor, we have: Plugging these values into the formula, we get:
Step 11:: Simplify the expression.
\frac{1}{\frac{1}{10} + \frac{1}{20} + \frac{1}{100} + \frac{1}{200}} = \frac{1}{\frac{20}{200} + \frac{10}{200} + \frac{2}{200} + \frac{1}{200}} = \frac{1}{\frac{33}{200}} = \frac{200}{33} \ \Omega
Final Answer
1. The total resistance of the first parallel circuit is $\frac{20}{3} \ \Omega$. 2. The total resistance of the second parallel circuit is $\frac{20}{3} \ \Omega$. 3. The total resistance of the third parallel circuit is $\frac{60}{7} \ \Omega$. 4. The total resistance of the fourth parallel circuit is 25 $\Omega$. 5. The total resistance of the fifth parallel circuit is $\frac{200}{33} \ \Omega$.
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