Math Problem Statement

Sebuah baterai dihubungkan dengan resistor akan menghasilkan arus 0,6 A. Jika pada rangkaian ditambahkan sebuah resistor 4,0 ohm yang dihubungkan seri dengan resistor pertama maka arus akan turun menjadi 0,5 A. Gaya gerak listrik (ggl) baterai (dalam volt) adalah .... a. 4 b. 5 c. 6 d. 12 e. 24

Solution

This problem is asking us to calculate the electromotive force (EMF or "gaya gerak listrik") of a battery in volts, using information about current and resistance.

Given dataThis problem is asking us to calculate the electromotive force (EMF or "gaya gerak listrik") of a battery in volts, using information about current and resistance.

Given

  • Initial current, I1=0.6AI_1 = 0.6 \, \text{A}
  • Current after adding a second resistor, I2=0.5AI_2 = 0.5 \, \text{A}
  • Resistance of the second resistor, R2=4.0ΩR_2 = 4.0 \, \Omega

The resistors are connected in series, which means their total resistance increases when the second resistor is added.

Step 1: Ohm's Law

We can use Ohm’s law: V=I×RV = I \times R Where:

  • VV is the electromotive force (EMF) or voltage
  • II is the current
  • RR is the resistance

Step 2: Determine the resistance of the first resistor

From the initial condition where the current is 0.6A0.6 \, \text{A}, we can assume that there is one resistor (call it R1R_1) connected to the battery.

Using Ohm’s law, the EMF of the battery is: V=I1×R1=0.6×R1V = I_1 \times R_1 = 0.6 \times R_1 This equation represents the total voltage or EMF.

Step 3: Analyze the new current after adding the second resistor

When the second resistor R2=4.0ΩR_2 = 4.0 \, \Omega is added in series, the total resistance becomes Rtotal=R1+4.0ΩR_{\text{total}} = R_1 + 4.0 \, \Omega, and the current drops to 0.5A0.5 \, \text{A}.

Using Ohm’s law again: V=I2×Rtotal=0.5×(R1+4.0)V = I_2 \times R_{\text{total}} = 0.5 \times (R_1 + 4.0) This represents the same EMF as in the first case.

Step 4: Equate the two expressions for voltage

From both scenarios, we know the voltage must be the same: 0.6×R1=0.5×(R1+4.0)0.6 \times R_1 = 0.5 \times (R_1 + 4.0)

Step 5: Solve for R1R_1

Expand the equation: 0.6R1=0.5R1+20.6 R_1 = 0.5 R_1 + 2 Now, subtract 0.5R10.5 R_1 from both sides: 0.1R1=20.1 R_1 = 2 Solve for R1R_1: R1=20.1=20ΩR_1 = \frac{2}{0.1} = 20 \, \Omega

Step 6: Calculate the EMF

Now that we know R1=20ΩR_1 = 20 \, \Omega, we can substitute it back into the equation for the EMF: V=0.6×20=12VV = 0.6 \times 20 = 12 \, \text{V}

Final Answer:

The EMF of the battery is 12 volts. Hence, the correct option is:

d. 12


Do you want further explanation or details on the steps? Here are some related questions for practice:

  1. How would the current change if the second resistor was 6 ohms instead of 4 ohms?
  2. What happens to the total resistance when resistors are added in series?
  3. How is the voltage across each resistor affected in a series circuit?
  4. What would be the new current if the second resistor was connected in parallel with the first?
  5. Can you derive the formula for total resistance in a parallel circuit?

Tip: When resistors are added in series, the total resistance increases, reducing the current if the voltage remains constant.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Ohm's Law
Series Circuit

Formulas

Ohm's Law: V = I × R
Series Circuit Total Resistance: R_total = R1 + R2

Theorems

Ohm's Law

Suitable Grade Level

Grades 10-12