1. How much power is dissipated by a 2.4 k resistor in series with a 12V power s

Question

1.     How much power is dissipated by a 2.4 k resistor in series with a 12V power supply?

2.     A series circuit contains four resistors with values of 4.7k, 5.6k, 8.1k and 10k. Which resistor has the smallest voltage drop?

3.     What is the current flow through R2 in the circuit shown below?

The above figure consists of schematic diagram of three resistors connected in series with a DC power supply of 36 volts with a return path to the negative terminal of the power supply also connected to ground.  

The first resistor is labeled

How much power is dissipated by a 2.4 k resistor in series with a 12V power supply? A series circuit contains four resistors with values of 4.7k, 5.6k, 8.1k and 10k. Which resistor has the smallest voltage drop? What is the current flow through R2 in the circuit shown below?

Explanation / Answer

V = 12 V
R = 2.4K Ohm = 2.4 x 10^3 Ohm

P = V^2 / R = 12^2/(2.4 x 10^3) = 144/2400 = 0.06 Watts

2.P=ExI=ExE/R=E^2/R

P=E^2/R 0.75W=E^2/4700 ohm E^2=0.75x4700 ohm E^2=3525 E=59.3 V

3.The reason we cant access the website is because its a https website a secured connection which makes me think that its a university website.

Try draw the circuit on here :D?

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4.The voltage between ground and the midpoint is a fraction of the full voltage depending on the fraction of the full resistance. I am assuming that R1 and R2 and R3 are all in series across the 36 volts. Let's say that R1 is at the top and R3 is at the bottom and that R2 is between R1 and R3. The total resistance is the sum of the three resistors so it is 4+8+12 = 24 Ohms. Between the R1 to R2 junction and the bottom there is the resistance of R2 +R3 = 8+12 = 20 Ohms. The fraction of the resistance at the R2 junction is 12/20 = .6 and that means that 0.6 of the 36 volts will be from there to the ground. So 0.6 x 36 = 21.6 volts is the answer. Draw it out on paper it it will be very clear.

A - Simply use the ratio of voltages to calculate the ratios of resistances.

B - Calculate the current in the series string from the voltage and resistance of the known resistor using: I = E/R Then calculalte the other resistors using R = E/I


A - 47 * 25/10 = 18.8

The other answer has the other (current) method "B".

6.Simplifying the above information, we know that:

VT = 12V (by VT I mean Total Voltage)
R1 = 220?
V2 = 7V
V3 = 2.5V

Since this is a series circuit and the only source of voltage is the 12V supply, we know from Kirchoff's voltage law that the total voltage in the loop must equal 0. Thus, the sum of the voltage DROPS must equal 12V. That is, V1 + V2 + V3 = VT. Substituting the known information in leads to V1 + 7V + 2.5V = 12V. Clearly, V1 must have a voltage drop of 2.5V.

Thus, V1 = 2.5V.

Now, since R1 is the only resistance known, and since we now know the voltage drop across it, we can calculate the current flowing across it, which is also the current through the entire circuit, from Ohm's Law. That is, I = V1/R1 Again substituting the known information, this leads to:

I = 2.5V/220? = 0.01136A = 11.36mA

Finally, we can calculate the unknown resistance values, R2 and R3, by using Ohm's Law with our now known current value.

R2 = V2/I
R2 = 7V/11.36mA = 616?

Likewise, we can go through the math to calculate R3 or simply note that V3 = V1, so R3 must be equal to R1, which is 220?.

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