Two 9.1 ohms resistors are connected in parallel, as are two 5.9 ohms resistors. These two combinations are then connected in series in a circuit with a 20 V battery.

What is the current in each resistor?
I9.1, I5.9 = A
What is the voltage across each resistor?
V9.1,V5.9 = V

To find the current in each resistor, we can use Ohm's Law, which states that the current flowing through a resistor is equal to the voltage across the resistor divided by the resistance.

First, let's find the equivalent resistance for the parallel combination of two 9.1 ohms resistors. In a parallel combination, the equivalent resistance (Rp) is given by the formula:
1/Rp = 1/R1 + 1/R2

Substituting the values, we have:
1/Rp = 1/9.1 + 1/9.1
1/Rp = 2/9.1
Rp = 9.1/2 = 4.55 ohms

Similarly, for the parallel combination of the two 5.9 ohms resistors, the equivalent resistance (Rq) is:
1/Rq = 1/R3 + 1/R4
1/Rq = 1/5.9 + 1/5.9
1/Rq = 2/5.9
Rq = 5.9/2 = 2.95 ohms

Now, the two combinations are connected in series, so their equivalent resistance (Rs) is simply the sum of the two individual resistances:
Rs = Rp + Rq
Rs = 4.55 + 2.95
Rs = 7.5 ohms

Now, we can calculate the total current (It) in the circuit using Ohm's Law:
It = Vt / Rt
It = 20 V / 7.5 ohms
It = 2.67 A

Since the combinations are connected in series, the current through each resistor will be the same as the total current. Therefore:
I9.1 = It = 2.67 A
I5.9 = It = 2.67 A

To find the voltage across each resistor, we need to apply Ohm's Law again. Since the current is the same through each resistor, we can use the formula V = I * R.

For the 9.1 ohms resistor:
V9.1 = I9.1 * R9.1
V9.1 = 2.67 A * 9.1 ohms
V9.1 = 24.17 V

For the 5.9 ohms resistor:
V5.9 = I5.9 * R5.9
V5.9 = 2.67 A * 5.9 ohms
V5.9 = 15.75 V

So, the current through each resistor is 2.67 A, and the voltage across the 9.1 ohms resistor is 24.17 V, while the voltage across the 5.9 ohms resistor is 15.75 V.

To find the current in each resistor, we can use Ohm's law which states that current (I) is equal to voltage (V) divided by resistance (R).

For the two 9.1 ohm resistors in parallel, we can first calculate the equivalent resistance using the formula for resistors in parallel:

1/Req = 1/R1 + 1/R2

1/Req = 1/9.1 + 1/9.1
= 2/9.1

Req = 9.1/2
= 4.55 ohms

Now, using Ohm's law, we can find the current flowing through the equivalent resistor:

I9.1 = V / Req
= 20 V / 4.55 ohms
≈ 4.40 A

For the two 5.9 ohm resistors in parallel, we can use the same process. First, calculate the equivalent resistance:

1/Req = 1/R1 + 1/R2

1/Req = 1/5.9 + 1/5.9
= 2/5.9

Req = 5.9/2
= 2.95 ohms

Now, using Ohm's law, we can find the current flowing through the equivalent resistor:

I5.9 = V / Req
= 20 V / 2.95 ohms
≈ 6.78 A

Therefore, the currents in the two 9.1 ohm resistors are approximately 4.40 A each, and the currents in the two 5.9 ohm resistors are approximately 6.78 A each.

To find the voltage across each resistor, we need to consider the series connection. In a series circuit, the total voltage across all the resistors is equal to the sum of the voltage drops across each resistor.

Since the resistors are connected in series, the current flowing through all the resistors is the same. Therefore, the voltage across each resistor can be calculated using Ohm's law:

V9.1 = I9.1 * R9.1
≈ 4.40 A * 9.1 ohms
≈ 40 V

V5.9 = I5.9 * R5.9
≈ 6.78 A * 5.9 ohms
≈ 40 V

Therefore, the voltage across each of the two 9.1 ohm resistors and the voltage across each of the two 5.9 ohm resistors is approximately 40 V.