Question

circuits lab
introduction: in this lab we will use ohms law to predict values for a circuit. then we will use a multimeter to measure an actual circuit to confirm our calculations.
show all work for calculations. submitting answers without work will result in zero credit.
calculating current
at the test station you will test the current at points a and b on the first circuit shown below. then you will remove one of the 5ω resistors and determine the current at point c.
original resistor removed

1. what will the current be at point a in the original circuit?

i = v/r 5 + 5+28.7 = 38.7ω

1. what will the current be at point b in the original circuit?

i = v/r v = 2.72v i = 2.72/38.7 = 0.07a

1. what will the current be at point c in the circuit with the resistor?
2. how will the brightness of the light - bulb be affected by removing a resistor?

Explanation:

Step1: Calculate total resistance for original circuit

For the original circuit, the resistors are in series. The total resistance $R_{total1}=5 + 5+28.7=38.7\Omega$.

Step2: Calculate current at point A in original circuit

Using Ohm's law $I = \frac{V}{R}$, with $V = 2.72V$ and $R = R_{total1}=38.7\Omega$, we have $I_A=\frac{2.72}{38.7}\approx0.07A$. In a series - circuit, the current is the same throughout, so the current at point A and point B in the original circuit is the same.

Step3: Calculate total resistance for circuit with resistor removed

When one of the 5 - $\Omega$ resistors is removed, the new total resistance $R_{total2}=5 + 28.7=33.7\Omega$.

Step4: Calculate current at point C in circuit with resistor removed

Using Ohm's law $I=\frac{V}{R}$, with $V = 2.72V$ and $R = R_{total2}=33.7\Omega$, we get $I_C=\frac{2.72}{33.7}\approx0.08A$.

Step5: Analyze effect on light - bulb brightness

The power of the light - bulb is given by $P = I^{2}R$ (assuming the resistance of the light - bulb $R$ is constant). When a resistor is removed, the total resistance of the circuit decreases, and according to Ohm's law $I=\frac{V}{R}$, the current in the circuit increases. Since $P = I^{2}R$, an increase in current will lead to an increase in power, and thus the brightness of the light - bulb will increase.

Answer:

1. $I_A\approx0.07A$
2. $I_B\approx0.07A$
3. $I_C\approx0.08A$
4. The brightness of the light - bulb will increase.