The circuit contains a “black box,” which you know conceals either an ideal (no

Question

The circuit contains a “black box,” which you know conceals either an ideal (no internal resistance) battery or a resistor, but you don’t know which. You read the current through the ammeter to be 2.8 A to the left.

a. Find the value that is read from the voltmeter, and indicate which voltmeter lead (top or bottom) is at the higher potential. Solve using a diffrent method than the one given.

One Possible Solution: We are given the voltage of the battery and the current through the 5 resistor, so using Kirchhoff’s loop rule around the outside loop gives us the current through the 10 resistor. Labeling the currents as shown above, we have:

outer loop, Clockwise: +12 V - (2.8 A)(5 ) - I2 (10) = 0 I2 = 0 .2 A

The negative sign indicates that the current is going in the direction opposite to how it is labeled. With the current going up in that branch, the bottom lead of the voltmeter must be at a higher potential than the top lead (current flows from high potential to low potential). The voltmeter reads the voltage drop across the resistor, which is the current times the resistance, or 2 V. [For the alternate solution, don’t compute I2 at all.]

b. Determine what the mystery component is. If the component is a battery, determine which is its positive lead (top or bottom), and its emf. If it is a resistor, find its resistance. Solve using a diffrent method than the one given.

One Possible Solution: We can first determine the current through the central branch from Kirchhoff’s junction rule. Note that the direction labeled for I2 is opposite to the actual current flow, so we need a minus sign when we compute the current from the voltage drop and resistance:

2.8 A = I1+I2   I1= 2.8 A I2 = 2.8 A (2 V)(10 ) = 3.0A

The positive sign indicates that the current is going downward in that branch, as it is drawn. Now use the loop rule for the left loop to determine the voltage change through the mystery element:

Left Loop, Clockwise: 12 V - (2.8 A)(5 ) - (3 A)(25 ) + V? = 0    V? = +77 V

[For the alternate solution, you may use the values for I1 and I2 computed above, but not Kirchhoff ’s rules.]

Explanation / Answer

the given possible solutions are correct:

you should proceed accordingly

(a)

We are given the voltage of the battery and the current through the 5 resistor, so using Kirchhoff’s loop rule around the outside loop gives us the current through the 10 resistor. Labeling the currents as shown above, we have:

outer loop, Clockwise: +12 V - (2.8 A)(5 ) - I2 (10) = 0 I2 = 0 .2 A

The negative sign indicates that the current is going in the direction opposite to how it is labeled. With the current going up in that branch, the bottom lead of the voltmeter must be at a higher potential than the top lead (current flows from high potential to low potential). The voltmeter reads the voltage drop across the resistor, which is the current times the resistance, or 2 V

(b)

We can first determine the current through the central branch from Kirchhoff’s junction rule. Note that the direction labeled for I2 is opposite to the actual current flow, so we need a minus sign when we compute the current from the voltage drop and resistance:

2.8 A = I1+I2   I1= 2.8 A I2 = 2.8 A (2 V)(10 ) = 3.0A

The positive sign indicates that the current is going downward in that branch, as it is drawn. Now use the loop rule for the left loop to determine the voltage change through the mystery element:

Left Loop, Clockwise: 12 V - (2.8 A)(5 ) - (3 A)(25 ) + V? = 0    V? = +77 V

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