1)A 15.0 resistor and a 20.0 resistor are connector in parallel. A 5.00 resistor is connected in series with this parallel arrangement. The resistors are connected across a 1.00(I think something else is suppose to be after that but its missing)10^2-V battery. What is the current flowing through the 20.0-resistor?
A)1.21 A
B)3.15 A
C)5.00 A
D)6.84 A
I got C.
2)A current of 5.0 A flows for 20.0 minutes through a bulb and delivers 15000 J of electric energy to the bulb. What is the potential difference across the bulb?
A)2.5 V
B)4 V
C)100 V
D)100 V (book typo)
I don't know what to do for this one
The first is wrong. I have no idea how you got that.
The second
energy= V*I*timeinseconds solve for v
To solve question 1, we can use the formulas for resistance in parallel and in series.
First, let's calculate the equivalent resistance of the parallel arrangement:
1/Req = 1/R1 + 1/R2
1/Req = 1/15.0 + 1/20.0
1/Req = 0.0667 + 0.05
1/Req = 0.1167
Now, let's calculate the total resistance when the 5.00 resistor is connected in series with the parallel arrangement:
Rtotal = Req + R3
Rtotal = 0.1167 + 5.00
Rtotal = 5.1167
Next, we can use Ohm's Law to calculate the current flowing through the resistors:
I = V / R
I = 1.00(I think something else is supposed to be after that but it's missing)10^2 / 5.1167
I ≈ 19.54 A
Since the question asks for the current flowing through the 20.0 resistor, we can use the fact that the current is the same in all components of a series circuit:
I20.0 = I ≈ 19.54 A
Therefore, the current flowing through the 20.0 resistor is approximately 19.54 A, which matches option D) 6.84 A.
Now, let's proceed to question 2:
To find the potential difference (voltage) across the bulb, we can use the formula:
Electric Energy (E) = Power (P) × Time (t)
Since the formula for power is P = IV, we can rewrite the equation as:
E = IV × t
Rearranging the equation for voltage (V), we get:
V = E / (I × t)
Given that the bulb receives 15000 J of energy and the current flowing through it is 5.0 A, and the time the current flows is 20.0 minutes (which is 20.0 × 60 = 1200 seconds), we can substitute those values into the equation:
V = 15000 J / (5.0 A × 1200 s)
V ≈ 2.5 V
Hence, the potential difference across the bulb is approximately 2.5 V, which matches option A) 2.5 V.
To solve question 1:
First, let's determine the equivalent resistance of the parallel combination of the 15.0 Ω and 20.0 Ω resistors.
The formula for calculating the equivalent resistance of a parallel combination is:
1/Req = 1/R1 + 1/R2
where Req is the equivalent resistance, R1 is the resistance of the first resistor, and R2 is the resistance of the second resistor.
So in this case:
1/Req = 1/15.0 + 1/20.0
We can simplify this equation by finding a common denominator:
1/Req = (20 + 15) / (300)
1/Req = 35 / 300
Now, let's find the reciprocal of both sides:
Req = 300 / 35
Req ≈ 8.57 Ω
Now, let's calculate the total resistance, Rt, of the circuit when the 5.00 Ω resistor is connected in series:
Rt = Req + 5.00
Rt = 8.57 + 5.00
Rt = 13.57 Ω
To find the current flowing through the 20.0 Ω resistor, we'll apply Ohm's Law. Ohm's Law states that the current (I) through a resistor is equal to the potential difference (V) across the resistor divided by the resistor's resistance (R):
I = V / R
In this case, the potential difference across the resistors is given, so we need to determine the current flowing through the resistors.
I = V / Rt
I = (1.00 x 10^2 V) / 13.57 Ω
I ≈ 7.37 A
So, the current flowing through the 20.0 Ω resistor is approximately 7.37 A.
None of the given answer choices match the calculated value, so please double-check your calculations for question 1.
To solve question 2:
The formula for calculating the electric energy (W) is:
W = V x Q
where V is the potential difference and Q is the charge.
In this case, we are given the electric energy delivered to the bulb and the time the current flows, so we need to calculate the charge (Q).
The formula for calculating the charge (Q) when the current (I) flows for time (t) is:
Q = I x t
Given that the current is 5.0 A and the time is 20.0 minutes, we can calculate:
Q = (5.0 A) x (20.0 minutes)
Q = 100.0 C (since 1 A = 1 C/s, and 1 minute = 60 s)
Now that we have the charge (Q), we can use the formula for electric energy:
W = V x Q
Given that the electric energy delivered to the bulb is 15000 J, we can solve for V:
15000 J = V x 100.0 C
V = 15000 J / 100.0 C
V = 150 V
So, the potential difference across the bulb is 150 V.