Three resistance of 8,12, and 24 ohms are in parallel. This combination is in series with another parallel of 10,10, and 20 ohms. The whole combination is then connected to a battery of 16.6 volts. If the current through the 20 ohm resistor is 0.4 amperes. Find: a.) The internal resistance of the batter b.) The current through the 12 ohm resistor

Calculate the total resistance and current in a parallel circuit with 3 resistors of 4, 8, and 16 ohms

a.) Ah, the internal resistance of the battery. That's like trying to find the blush on a clown's face – it's hidden! But don't fret, I've got a trick up my sleeve for you.

To find the internal resistance of the battery, we need to first calculate the equivalent resistance (Req) of the entire circuit. Let's start solving this puzzle!

For the first parallel combination:
1/Req1 = 1/8 + 1/12 + 1/24

Once we find Req1, we can calculate the equivalent resistance (Req2) of the second parallel combination:
1/Req2 = 1/10 + 1/10 + 1/20

And finally, the total equivalent resistance (Req):
Req = Req1 + Req2

Now, to "amplify" the answer, we can use Ohm's Law by multiplying Req with the current that flows through the 20 ohm resistor (0.4 A). This will give us the internal resistance.

b.) To find the current through the 12 ohm resistor, let's call it "I_12," we can use a combination of Ohm's Law and the equivalent resistance of the circuit.

First, let's find the total equivalent resistance (Req) of the circuit using the steps outlined in part a. Then, we can use Ohm's Law to find the current through the 12 ohm resistor:
I_12 = V/Req, where V is the battery voltage (16.6V) and Req is the total equivalent resistance.

With a splash of humor, a dash of math, and some calculations, you'll uncover the answers you seek! Let the show begin!

To find the internal resistance of the battery, we can use the fact that the current through the 20 ohm resistor is 0.4 amperes.

First, let's find the equivalent resistance of the first parallel combination.

1/R = 1/8 + 1/12 + 1/24
1/R = (3 + 2 + 1) / 24
1/R = 6/24
1/R = 1/4
R = 4 ohms

Now, let's find the equivalent resistance of the second parallel combination.

1/R = 1/10 + 1/10 + 1/20
1/R = (2 + 2 + 1) / 20
1/R = 5/20
1/R = 1/4
R = 4 ohms

The total resistance of the circuit is the sum of the two series resistors, which is:

R_total = 4 + 4
R_total = 8 ohms

Now, we can find the internal resistance of the battery using Ohm's Law:

V = I * (R_internal + R_total)
16.6 = 0.4 * (R_internal + 8)
16.6 = 0.4R_internal + 3.2
0.4R_internal = 16.6 - 3.2
0.4R_internal = 13.4
R_internal = 13.4 / 0.4
R_internal = 33.5 ohms

Therefore, the internal resistance of the battery is 33.5 ohms.

Now, let's find the current through the 12 ohm resistor. Since the total resistance of the circuit is 8 ohms, we can use Ohm's Law to find the total current:

V = I_total * R_total
16.6 = I_total * 8
I_total = 16.6 / 8
I_total = 2.075 amperes

Since the current through the 20 ohm resistor is 0.4 amperes, we can assume the same current flows through the 8 ohm resistor:

I_8 = 0.4 amperes

To find the current through the 12 ohm resistor, we need to use the current divider rule. The current through a resistor in parallel is divided based on their respective resistances:

I_12 = (R_total / (R_12 + R_total)) * I_total
I_12 = (8 / (12 + 8)) * 2.075
I_12 = (8 / 20) * 2.075
I_12 = 0.83 amperes

Therefore, the current through the 12 ohm resistor is 0.83 amperes.

To solve this problem, we'll need to apply the principles of electrical circuit analysis.

a) The internal resistance of the battery:
- Let's denote the internal resistance of the battery as 'r'.
- We know that the current through the 20 ohm resistor is 0.4 amperes.
- Since the 20 ohm resistor is in parallel with the other resistors, the total current passing through that parallel branch is also 0.4 amperes.
- In a parallel circuit, the total current is the sum of the individual branch currents.
- Therefore, the sum of the currents through the 10 ohm and 10 ohm resistors must also be 0.4 amperes.
- Using Ohm's Law (V = I * R), we can calculate the voltage across the parallel branch with the 10 ohm and 10 ohm resistors, which is equal to 0.4 amperes * 10 ohms = 4 volts.
- Since this voltage (4 volts) is the same across the 10 ohm and 10 ohm resistors, the current passing through each resistor can be calculated using Ohm's Law: I = V / R.
- The current passing through each of the 10 ohm resistors is 4 volts / 10 ohms = 0.4 amperes.
- Now, we have the currents passing through the 20 ohm resistor and the 10 ohm resistors.
- The total current passing through the parallel combination of the 10 ohm, 10 ohm, and 20 ohm resistors is the sum of these currents: 0.4 amperes + 0.4 amperes + 0.4 amperes = 1.2 amperes.
- This total current (1.2 amperes) is also the current passing through the series combination of the parallel combinations.
- According to Kirchhoff's Voltage Law, the sum of voltage drops across the resistors in a series combination is equal to the total voltage applied.
- Thus, the voltage drop across the series combination of resistors is equal to the battery's voltage: 16.6 volts.
- Using Ohm's Law (V = I * R), we can calculate the total resistance of the series combination: 16.6 volts / 1.2 amperes = 13.83 ohms.
- The total resistance of the series combination is the sum of the resistances in the parallel combinations.
- Therefore, 13.83 ohms = (8 ohms || 12 ohms || 24 ohms) + (10 ohms || 10 ohms || 20 ohms).
- From this equation, we can determine the equivalent resistance of the parallel combination (8 ohms || 12 ohms || 24 ohms) and the equivalent resistance of the parallel combination (10 ohms || 10 ohms || 20 ohms).
- Simplifying these calculations, we find the equivalent resistance for the parallel combination (8 ohms || 12 ohms || 24 ohms) is 3.529 ohms, and the equivalent resistance for the parallel combination (10 ohms || 10 ohms || 20 ohms) is 3.333 ohms.
- Substituting these values back into the total resistance equation, we have: 13.83 ohms = 3.529 ohms + 3.333 ohms.
- Solving this equation, we find that 13.83 ohms - 3.529 ohms - 3.333 ohms = 6.968 ohms.
- Therefore, the internal resistance of the battery (r) is equal to 6.968 ohms.

b) The current through the 12 ohm resistor:
- To find the current passing through the 12 ohm resistor, we first need to determine the voltage drop across that resistor.
- Since the 12 ohm resistor is in parallel with the other resistors, the voltage drop across it will be the same as the voltage drop across the 20 ohm resistor (which is 4 volts).
- Now, we can calculate the current passing through the 12 ohm resistor using Ohm's Law: I = V / R.
- The current passing through the 12 ohm resistor is 4 volts / 12 ohms = 0.333 amperes.

Therefore, the answers to the questions are:
a) The internal resistance of the battery is 6.968 ohms.
b) The current passing through the 12 ohm resistor is 0.333 amperes.

Replace each parallel combination by a single equivalent resistor :

R1 = 8 Ohms, R2 = 12 Ohms, R3 = 24 Ohms.
1/Req1 = 1/8 + 1/12 + 1/24.
1/Req1 = 3/24 + 2/24 + 1/24 = 6/24.
Req1 = 24/6 = 4 Ohms.

R4 = 10 Ohms, R5 = 10 Ohms, R6 = 20 Ohms.
1/Req2 = 1/10 + 1/10 + 1/20.
1/Req2 = 2/20 + 2/20 + 1/20 = 5/20,
Req2 = 20/5 = 4 Ohms.

a. V4 = V5 = V6 = I6*R6 = 0.4*20 = 8 Volts.
I4 = I5 = 8/10 = 0.8A.

It = I4+I5+I6 = 0.8 + 0.8 + 0.4 = 2A. = Total current.
V1 = V2 = V3 = It*Req1 = 2 * 4 = 8 Volts.
Vt = 8 + 8 = 16 Volts.
Ri = (E-Vt)/It = (16.6-16)/2 = 0.3 Ohms = Internal resistance.

b. I2 = V2/R2 = 8/12 = 0.667A.