Two bulbs each of resistance 2.0 ohms are connected in parallel with a battery. Calculate the current flowing through any of the bulbs if the voltage of the battery is 0.2V
Please I need the total answer
parallel means 0.2V across 2.0Ω
current = voltage / resistance
The answer is 20v
Well, if I may interrupt for a moment, I have a joke about resistance that's quite electric.
Why did the electron go to jail?
Because it had too much resistance!
Okay, back to your question. To calculate the current flowing through any of the bulbs, you can use Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across it divided by the resistance (R).
In this case, the voltage of the battery is given as 0.2V. Since the two bulbs are connected in parallel, they have the same voltage across them (0.2V). Additionally, since the resistance of each bulb is 2.0 ohms, we can calculate the current flowing through any of the bulbs using the formula:
I = V / R
I = 0.2V / 2.0 ohms
I = 0.1 A
So, the current flowing through any of the bulbs would be 0.1 Amps. But let's hope they don't blow a fuse with their shocking behavior!
To calculate the current flowing through any of the bulbs, we can use Ohm's Law, which states that the current (I) flowing through a component is equal to the voltage (V) across the component divided by its resistance (R).
In this case, we have two bulbs with a resistance of 2.0 ohms each. Since the bulbs are connected in parallel, the total resistance (RT) is given by the formula:
1/RT = 1/R1 + 1/R2
Where R1 and R2 are the resistances of the individual bulbs. Substituting the values into the equation:
1/RT = 1/2.0 + 1/2.0
Simplifying the equation:
1/RT = 2/2.0
1/RT = 1.0
RT = 1.0 ohm
Now that we know the total resistance of the circuit, we can calculate the current (I) flowing through any of the bulbs using Ohm's Law:
I = V / RT
Substituting the given values into the equation:
I = 0.2V / 1.0 ohm
I = 0.2 A
Therefore, the current flowing through any of the bulbs is 0.2 Amperes (A).