A 37 ohm lamp and a 59 ohm lamp are connected in series and placed across a potential of 43 V.
What is the equivalent resistance of the
circuit?
What is the current in the circuit?
Answer in units of A.
What is the voltage drop across the first
lamp?
Answer in units of V.
What is the power dissipated in the first
lamp?
Answer in units of W.
Req. = R1+R2 = 37 + 59 = 96 Ohms.
I = E/Req. = 43/96 = 0.448A
V1 = I*R1 = 0.448 * 37 = 16.6 Volts.
P1 = V1*I = 16.6 * 0.448 = 7.44 W.
I will be happy to critique your thinking on this.
I was only able to get the first and second question. Not the last two.
Well, I'm no mathematician, but I can definitely clown around with some numbers! Let's get started, shall we?
To find the equivalent resistance of the circuit when the lamps are connected in series, we simply add the resistances. So the equivalent resistance will be 37 ohms + 59 ohms = 96 ohms. Easy peasy, right?
Now, to find the current in the circuit, we can use Ohm's Law: I = V / R. Plugging in the potential of 43 V and the equivalent resistance of 96 ohms, we get I = 43 V / 96 ohms = 0.4479... A. Let's just round that up to 0.448 A because, hey, it's a good enough approximation!
Now, the voltage drop across the first lamp will be the same as the potential (43 V) because the lamps are connected in series. So the voltage drop across the first lamp is 43 V. No funny business there!
Lastly, to find the power dissipated in the first lamp, we can use the formula P = I * V. Using the current of 0.448 A and the voltage drop of 43 V, we get P = 0.448 A * 43 V = 19.264 W. So the power dissipated in the first lamp is 19.264 W.
Voilà! The calculations are complete, and I've done my clown duty. I hope that brings a smile to your face!
To solve these questions, we need to use Ohm's Law and the formulas for series circuits. Let's break it down step by step.
1. Equivalent Resistance (Req):
In a series circuit, the total resistance is the sum of the individual resistances. So, to find the equivalent resistance of the circuit, simply add the resistances of the lamps: Req = 37 ohms + 59 ohms = 96 ohms.
2. Current (I):
Using Ohm's Law (V = IR), we can rearrange the formula to solve for current. In this case, the voltage (V) is given as 43V, and the equivalent resistance (Req) is 96 ohms. So, I = V / Req = 43V / 96 ohms = 0.447 A.
3. Voltage drop across the first lamp:
In a series circuit, the voltage is divided among the components based on their resistance. Since the lamps are in series, the total voltage across both lamps (43V) will be divided according to their resistance values. To find the voltage drop across the first lamp, calculate its share based on its resistance value:
Voltage drop across the first lamp = (Resistance of the first lamp / Total resistance) * Total voltage.
Here, the resistance of the first lamp is 37 ohms, and the total resistance is 96 ohms, so the voltage drop across the first lamp would be: (37 ohms / 96 ohms) * 43V = 16.66 V (rounded to two decimal places).
4. Power dissipated in the first lamp:
To find the power dissipated in the first lamp, we can use the formula P = IV, where P is power, I is current, and V is voltage. In this case, the current (I) is 0.447 A (calculated earlier), and the voltage drop across the first lamp (V) is 16.66V (calculated above). So, P = I * V = 0.447 A * 16.66 V = 7.44 W (rounded to two decimal places).
So, to summarize:
- The equivalent resistance of the circuit is 96 ohms.
- The current in the circuit is 0.447 A.
- The voltage drop across the first lamp is 16.66V.
- The power dissipated in the first lamp is 7.44 W.