QUESTION IMAGE
Question
a parallel circuit is constructed of three resistors. the first is 15 ω, the second is 25 ω and the third drops 12 v when 0.2 a flows through it.
second voltage drop = ____ v
third resistance = ____ ω
total voltage = ____ v
first voltage drop = ____ v
first current = ____ a
total resistance = ____ ω
second current = ____ a
total current = ____ a
Step1: Determine the voltage across each resistor
In a parallel circuit, the voltage across each resistor is the same. The third resistor has a voltage drop of \(12\ V\). So, the voltage across the first resistor (\(V_1\)), the second resistor (\(V_2\)) and the total voltage (\(V_{total}\)) is \(12\ V\).
Step2: Calculate the current through the first resistor
Using Ohm's law \(I=\frac{V}{R}\), for the first resistor \(R_1 = 15\ \Omega\) and \(V_1=12\ V\). Then \(I_1=\frac{V_1}{R_1}=\frac{12}{15}=0.8\ A\)
Step3: Calculate the current through the second resistor
For the second resistor \(R_2 = 25\ \Omega\) and \(V_2 = 12\ V\). Using Ohm's law \(I=\frac{V}{R}\), \(I_2=\frac{V_2}{R_2}=\frac{12}{25}=0.48\ A\)
Step4: Calculate the resistance of the third resistor
For the third resistor, given \(V_3 = 12\ V\) and \(I_3=0.2\ A\). Using Ohm's law \(R=\frac{V}{I}\), \(R_3=\frac{V_3}{I_3}=\frac{12}{0.2}=60\ \Omega\)
Step5: Calculate the total resistance
Using the formula for parallel resistors \(\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\). Substituting \(R_1 = 15\ \Omega\), \(R_2 = 25\ \Omega\) and \(R_3=60\ \Omega\)
Step6: Calculate the total current
Using Ohm's law \(I_{total}=\frac{V_{total}}{R_{total}}\), with \(V_{total} = 12\ V\) and \(R_{total}\approx8\ \Omega\), \(I_{total}=\frac{12}{8}=1.5\ A\) (but also using \(I_{total}=I_1 + I_2+I_3\), \(I_1 = 0.8\ A\), \(I_2=0.48\ A\), \(I_3 = 0.2\ A\), \(I_{total}=0.8 + 0.48+0.2=1.48\ A\))
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Second voltage drop = \(12\ V\)
Third resistance=\(60\ \Omega\)
Total voltage=\(12\ V\)
First voltage drop=\(12\ V\)
First current=\(0.8\ A\)
Total resistance=\(8\ \Omega\)
Second current=\(0.48\ A\)
Total current=\(1.48\ A\)