Question

Compute the equivalent resistance of the network in the figure below, and find the current in each resistor. (R1 = 4.00 Ω, R2 = 7.00 Ω, R3 = 10.0 Ω, R4 = 2.00 Ω, and ε = 61.0 V.) The battery has negligible internal resistance. equivalent resistance Incorrect: Your answer is incorrect. Ω current in each resistor Incorrect: Your answer is incorrect. A (Resistor R1) A (Resistor R2) A (Resistor R3) A (Resistor R4)

Answer 1

DATA:

$\\ \varepsilon =61.0\,V \\R_{1}=4.00\,\Omega \\R_{2}=7.00\,\Omega \\R_{3}=10.0\,\Omega\\R_{4}=2.00\,\Omega$

SOLUTION

Part a) Compute the equivalent resistance of the network

The resistors $R_{1}$ and $R_{3}$ are connected in Series (see figure), so:

Replacing data given values:

$R_{13}=4.00\,\Omega +10.0\,\Omega \, \, \, \, \, \, \, \, \, \,$

Solving we obtain:

${\color{Red} R_{13}=14.0\,\Omega }\, \, \, \, \, \, \, \, \, \, (2)$

Simillarly, resistors $R_{2}$ and $R_{4}$ are connected in Series, so:

Replacing data given values:

$R_{24}=7.00\,\Omega +2.00\,\Omega \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,$

Solving we obtain:

${\color{Red} R_{24}=9.00\,\Omega} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (4)$

Now, the circuit can be re-drawn as follows:

13 24

Now, resistors $R_{13}$ and $R_{24}$ are connected in Parallel, so, the equivalent resistance $R_{e}$ is:

$\frac{1}{R_{e}}=\frac{1}{R_{13}}+\frac{1}{R_{24}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (5)$

Replacing (2) and (4) into equation (5) we have:

$\frac{1}{R_{e}}=\frac{1}{14.0\,\Omega }+\frac{1}{9.00\,\Omega }\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,$

Solving we obtain that the equivalent resistance $R_{e}$ of the network is:

${\color{Blue} R_{e}=5.48\,\Omega }\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (6)$

Part b) find the current in each resistor​.

The circuit can be re-drawn as follows:

Therefore, the total current $I_{e}$ is:

$I_{e}=\frac{\varepsilon }{R_{e}}$

Replacing values:

$I_{e}=\frac{61.0\,V}{5.48\,\Omega }$

Solving we obtain:

But, the equivalent resistance comes from parallel connection, so:

$\varepsilon =V_{13}=V_{24}=61.0\,V$

Therefore:

$I_{13}=\frac{V_{13}}{R_{13}}$

Replacing values:

$I_{13}=\frac{61.0\,V}{14.0\,\Omega }$

Solving we obtain:

$I_{13}=4.36\,A$

But, the resistor $R_{13}$ comes from a Series connection, so:

$I_{13}=I_{1}=I_{3}=4.36\,A$

Therefore:

$\\ {\color{Blue} I_{1}=4.36\,A}\, \, \,\, \, \, \, \, \, \, \, and\, \, \, \\ {\color{Blue} I_{3}=4.36\,A}$

Simillarly,

$I_{24}=\frac{V_{24}}{R_{24}}$

Replaicing values:

$I_{24}=\frac{61.0\,V}{9.00\,\Omega }$

Solving we obtain:

$I_{24}=6.78\,A$

But, $R_{24}$ comes from a series connection:

$I_{24}=I_{2}=I_{4}=6.78\,A$

Finally, the current in each resistor​ is:

$\\ {\color{Blue} I_{1}=4.36\,A} \\ {\color{Red} I_{2}=6.78\,A} \\ {\color{Green} I_{3}=4.36\,A}\\ {\color{Cyan} I_{4}=6.78\,A}$