Physics, 13.12.2019 05:31 marie0321rivera67

Pz a hair dryer with a resistance of 12.0 w and a lamp with a resistance of 125 w are connected in parallel to a 125-v source through a 1.50-w resistor in a series. find the current through the lamp when the hair dryer is on.

0.880 a

8.80 a

88.0 a

880 a

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Answer from: momodagrate

the question is not complete, below is the complete question

A hair dryer with a resistance of 12.0 Ω and a lamp with a resistance of 125 Ω are connected in parallel to a 125-V source through a 1.50- Ω resistor in series. Find the current through the lamp when the hair dryer is on.

0.88A

Explanation:

The representation of the circuit is shown in the attached file where the V represent the voltage source,the $R_{a} ,R_{b} ,R_{c}$ represent the 1.50Ω,12Ω and 125Ω, resistor respectively.

We have to model the circuit into a single voltage source and a single resistance. This can be done by combining the resistance value.

First, the hair dryer resistance and the lamp resistance are arranged in parallel.The equivalent resistance for the two component is determine below

$=\frac{R_{b}*R_{c} }{R_{b}+R_{c}} \\R_{bc}=\frac{12*125 }{12+125}\\$

$R_{bc}=10.9 ohms\\$ .

Now the $R_{bc}=10.9 ohms$ . is in series with $R_{a}\\$ .

the equivalent value of the new formed is resistor is determine below

$R_{abc}=R_{a}+R_{bc}=1.5+10.9\\R_{abc}=12.4ohms\\$

Now we have a single resistor and a voltage source of value 125v

Using homes law, we can determine the current that pass through $R_{abc}$ resistor

V=IR

$I=\frac{125}{12.4}\\I=10.1A$

Recall, when resistors are arrange in series the same amount of current pass through them, and when in parallel, different amount of current will pass through them depending on the value.

hence we can conclude that, 10.1A will pass through the $R_{bc}$ and $R_{a}$ resistor. We can now determine the current through the lamp when the hair dryer is on using the simple circuit in the second attachment. The expression for the current through the lamp is

$I_{c}=\frac{I*R_{b} }{R_{b}+R_{c}} \\$

when we substitute value, we arrive at

$I_{c}=\frac{10.1*12 }{12+125} \\I_{c}=0.88A\\$

hence the value of the current through the lamp when the dryer is on is 0.88A

Pz a hair dryer with a resistance of 12.0 w and a lamp with a resistance of 125 w are connected in

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