Two resistors of 2.0 ohm and 3.0 ohm are connected a) in series b) in parallel, with a battery of 6.0V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery

Question

Two resistors of 2.0 ohm and 3.0 ohm are connected a) in series b) in parallel, with a battery of 6.0V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery

Answer 1

I'm sure you have seen examples of this simple circuit.

(a) R = 2+3
(b) 1/R = 1/2 + 1/3

I = E/R in each case.

Answer 2

To solve this problem, we first need to understand how resistors behave when connected in series and parallel.

a) When resistors are connected in series, the total resistance (R_total) is the sum of the individual resistances (R1 + R2).

b) When resistors are connected in parallel, the reciprocal of the total resistance (1/R_total) is the sum of the reciprocals of the individual resistances (1/R1 + 1/R2).

Now let's solve the problem step by step:

a) When the resistors are connected in series:

Circuit diagram:
```
__________ ________ _______
| | | | | |
| Battery |---- Resistor 1 ---- Resistor 2 ----
|_________| |_______| |______|
```

Here, R1 = 2.0 ohm and R2 = 3.0 ohm.

1. Calculate the total resistance (R_total) by adding the individual resistances: R_total = R1 + R2 = 2.0 + 3.0 = 5.0 ohm.

2. Use Ohm's Law (V = I * R) to calculate the current (I) through the battery. Since the internal resistance is negligible, the voltage across the resistors is equal to the battery voltage. Thus, V = 6.0V.

I = V / R_total = 6.0V / 5.0 ohm = 1.2A

Therefore, the current through the battery in the series circuit is 1.2 Amperes.

b) When the resistors are connected in parallel:

Circuit diagram:
```
___________ ________
| | | |
| Battery |---- Resistor 1
|__________| |_______|
________
| |
| Resistor 2
|_______|
```

Here, R1 = 2.0 ohm and R2 = 3.0 ohm.

1. Calculate the reciprocal of the individual resistances: 1/R1 = 1/2.0 ohm and 1/R2 = 1/3.0 ohm.

2. Calculate the reciprocal of the total resistance (R_total) by adding the reciprocals of the individual resistances: 1/R_total = 1/R1 + 1/R2.

1/R_total = 1/2.0 + 1/3.0 = (3 + 2) / (2 * 3) = 5/6

R_total = 6/5 ohm (taking the reciprocal)

3. Use Ohm's Law (V = I * R) to calculate the current (I) through the battery. V = 6.0V (same as before).

I = V / R_total = 6.0V / (6/5) ohm = 5A

Therefore, the current through the battery in the parallel circuit is 5 Amperes.