A 12.0 V car battery with negligible internal resistance is connected to a series combination of a 3.5 ohms resistor that obeys Ohm's law and a thermistor that does not obey Ohm's law but instead has a current-voltage relationship V = 4(I) + 1.5(I^2)
What is the current through the 3.5 ohms resistor?
What is the current through the 3.5 ohms resistor?
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Since they are in series we can assume that the total voltage drop of each component is equal to the source and there is only one current flowing:
V1 + V2 = 12
where:
V1 = 3.5(I)
V2 = 1.5(I^2) + 4(I)
thus:
3.5(I) + 1.5(I^2) + 4(I) = 12
1.5(I^2) + 7.5(I) = 12
solving quadratic equation
I = 1.275 amps
V1 + V2 = 12
where:
V1 = 3.5(I)
V2 = 1.5(I^2) + 4(I)
thus:
3.5(I) + 1.5(I^2) + 4(I) = 12
1.5(I^2) + 7.5(I) = 12
solving quadratic equation
I = 1.275 amps
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Let the current through the resistor be 'I'.
Hence the voltage across the resistor = 3.5*I = 3.5I
Now the total voltage of 12 V must drop across the resistor and the thermistor, so that
Voltage across resistor + Voltage across the thermistor= 12.0
3.5I +4(I) + 1.5(I^2)=12
7.5 I+ 1.5(I^2) =12
1.5(I^2) + 7.5 I -12= 0
I^2 + 5 I -8=0 [dividing whole equation by '1.5']
I^2 + 5 I -8=0
I = 1.27 Amps or -6.27 Amps [solving quadratic equation]
Hence the voltage across the resistor = 3.5*I = 3.5I
Now the total voltage of 12 V must drop across the resistor and the thermistor, so that
Voltage across resistor + Voltage across the thermistor= 12.0
3.5I +4(I) + 1.5(I^2)=12
7.5 I+ 1.5(I^2) =12
1.5(I^2) + 7.5 I -12= 0
I^2 + 5 I -8=0 [dividing whole equation by '1.5']
I^2 + 5 I -8=0
I = 1.27 Amps or -6.27 Amps [solving quadratic equation]
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The resistor and thermistor are in series, so the current is the same.
Voltage drop across the resistor = IR = 3.5I
Voltage drop across the thermistor = 4(I) + 1.5(I^2)
Because they are in series, the voltage drop across the resistor PLUS the voltage drop across the thermistor equals 12V (Kirchhoff's second law).
3.5I + 4(I) + 1.5(I^2) = 12
7.5I + 1.5I^2 = 12
This is a quadratic equation: 1.5I^2 + 7.5I -12 = 0
Solving, using the standard quadratic formula gives:
I = [ -7.5 +/- sqrt(7.5^2 - 4x1.5x-12) ] / (2x1.5)
= [ -7.5 +/-sqrt (128.25) ] / 3
= [ -7.5 +/- (11.3) ] / 3
Consider only the positive solution:
I = [ -7.5 + 11.3 ] / 3
= 1.3A
Voltage drop across the resistor = IR = 3.5I
Voltage drop across the thermistor = 4(I) + 1.5(I^2)
Because they are in series, the voltage drop across the resistor PLUS the voltage drop across the thermistor equals 12V (Kirchhoff's second law).
3.5I + 4(I) + 1.5(I^2) = 12
7.5I + 1.5I^2 = 12
This is a quadratic equation: 1.5I^2 + 7.5I -12 = 0
Solving, using the standard quadratic formula gives:
I = [ -7.5 +/- sqrt(7.5^2 - 4x1.5x-12) ] / (2x1.5)
= [ -7.5 +/-sqrt (128.25) ] / 3
= [ -7.5 +/- (11.3) ] / 3
Consider only the positive solution:
I = [ -7.5 + 11.3 ] / 3
= 1.3A
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The same current flows through both the resistor and the thermistor.
We know from Kirchhoff's voltage law that
[voltage across resistor] + [voltage across thermistor] = 12.0v
So using Ohm's Law on the resistor v = ir
3.5(I) + 4(I) + 1.5(I²) = 12
1.5I² + 7.5I - 12 = 0
Solve the quadratic for I
I = 1.275A or I = -6.27A
It looks like the answer is 1.275A
Just to check:
Now use Ohm's law again on the resistor
v = ir
v = 1.275 * 3.5 = 4.46 V
or
v = -6.25 * 3.5 = Whatever that is, it's more than the battery voltage, so that can be discounted!
We know from Kirchhoff's voltage law that
[voltage across resistor] + [voltage across thermistor] = 12.0v
So using Ohm's Law on the resistor v = ir
3.5(I) + 4(I) + 1.5(I²) = 12
1.5I² + 7.5I - 12 = 0
Solve the quadratic for I
I = 1.275A or I = -6.27A
It looks like the answer is 1.275A
Just to check:
Now use Ohm's law again on the resistor
v = ir
v = 1.275 * 3.5 = 4.46 V
or
v = -6.25 * 3.5 = Whatever that is, it's more than the battery voltage, so that can be discounted!