Question

Please help with Physics Circuits!

Answer 1

1) Let's start by calculating the equivalent resistance of the three resistors in parallel, $R_2, R_3, R_4$:
$\frac{1}{R_{234}}= \frac{1}{R_2}+ \frac{1}{R_3}+ \frac{1}{R_4}= \frac{1}{4.5 \Omega}+ \frac{1}{1.3 \Omega}+ \frac{1}{6.3 \Omega}=1.15 \Omega^{-1}$
From which we find
$R_{234}= \frac{1}{1.15 \Omega^{-1}}=0.9 \Omega$

Now all the resistors are in series, so the equivalent resistance of the circuit is the sum of all the resistances:
$R_{eq}=R_1 + R_{234} = 5 \Omega + 0.9 \Omega = 5.9 \Omega$
So, the correct answer is D) 


2) Let's start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}$
From which we find
$R_{23} = 2.5 \Omega$

And these are connected in series with a resistor of $10 \Omega$, so the equivalent resistance of the circuit is
$R_{eq}=10 \Omega + 2.5 \Omega = 12.5 \Omega$

And by using Ohm's law we find the current in the circuit:
$I= \frac{V}{R_{eq}}= \frac{9 V}{12.5 \Omega}=0.72 A$
So, the correct answer is C).


3) Let' start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}$
From which we find
$R_{23} = 2.5 \Omega$
Then these are in series with all the other resistors, so the equivalent resistance of the circuit is
$R_{eq}=R_1 + R_{23}+R_4 = 5 \Omega + 2.5 \Omega + 5 \Omega =12.5 \Omega$

And by using Ohm's law we find the current flowing in the circuit:
$I= \frac{V}{R_{eq}}= \frac{12 V}{12.5 \Omega}=0.96 A$

And so the voltage read by the voltmeter V1 is the voltage drop across the resistor 2-3:
$V= I R_{23} = (0.96 A)(2.5 \Omega)=2.4 V$
So, the correct answer is D).


4) Again, let's start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{13 \Omega}+ \frac{1}{18 \Omega}=0.13 \Omega^{-1}$
From which we find
$R_{23} = 7.55 \Omega$

Now all the resistors are in series, so the equivalent resistance of the circuit is:
$R_{eq}= R_1 + R_{23}+R_4=8.5 \Omega+7.55 \Omega + 3.2 \Omega = 19.25 \Omega$

The current in the circuit is given by Ohm's law
$I= \frac{V}{R_{Eq}}= \frac{15 V}{19.25 \Omega}=0.78 A$

Now we can compare the voltage drops across the resistors. Resistor 1:
$V_1 = I R_1 = (0.78 A)(8.5 \Omega)=6.63 V$
Resistor 2 and resistor 3 are in parallel, so they have the same voltage drop:
$V_2 = V_3 = V_{23} = I R_{23} = (0.78 A)(7.55 \Omega)=5.89 V$
Resistor 4:
$V_4 = I R_4 = (0.78 A)(3.2 \Omega)=2.50 V$

So, the greatest voltage drop is on resistor 1, so the correct answer is D).


5) the figure shows a circuit with a resistor R and a capacitor C, so it is an example of RC circuit. Therefore, the correct answer is D).

6) The circuit is the same as part 4), so the calculations are exactly the same. Therefore, the power dissipated on resistor 3 is
$P_3 = I_3^2 R_3 = \frac{V_3^2}{R_3}= \frac{(5.89 V)^2}{18 \Omega}=2.0 W$
So, correct answer is B).

7) The circuit is the same as part 4), so we can use exactly the same calculation, and we immediately see that the resistor with lowest voltage drop was R4 (2.50 V), so the correct answer is B) R4.

Answer 2

Answer:

The person that answered before me is right except for question 1

1.    5.9 Ω

2.   0.72 A

3.   2.4 V

4.   R1

5.   RC circuit

6.   2.0 W

7.   R4

Explanation:

I took the test and I got question 1 wrong. (K12 Physics sem 2 8.05)