A simple circuit contains an ideal battery and a resistance R. If a second resistor is placed in parallel with the first.

Options

A. the potential across R will decrease
B. the current through R will decreased
C. the current delivered by the battery will increase
D. the power dissipated by R will increased.

Answer

To solve the problem of a simple circuit containing an ideal battery and a resistance R, with a second resistor placed in parallel with the first, we will analyze the effects of this configuration step by step.

### Step 1: Understand the Circuit Configuration
In the original circuit, we have an ideal battery with an EMF (E) connected to a resistor R. When a second resistor (let's call it R2) is placed in parallel with R, we need to analyze how this affects the overall circuit.

**Hint:** Remember that resistors in parallel share the same voltage across them, which is equal to the EMF of the battery.

### Step 2: Calculate the Equivalent Resistance
The equivalent resistance (R_eq) of two resistors in parallel (R and R2) can be calculated using the formula:
\[ \frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{R2} \]
This can be rearranged to find R_eq:
\[ R_{eq} = \frac{R \cdot R2}{R + R2} \]

**Hint:** If R2 is equal to R, then the equivalent resistance becomes R/2.

### Step 3: Analyze the Voltage Across Resistors
Since R and R2 are in parallel, the voltage across both resistors remains equal to the EMF of the battery (E). Thus, the voltage across R does not change.

**Hint:** The voltage across resistors in parallel is always the same and equal to the voltage of the source.

### Step 4: Calculate the Current Through Each Resistor
Using Ohm's Law (V = IR), the current through resistor R (I_R) can be calculated as:
\[ I_R = \frac{E}{R} \]
Similarly, the current through resistor R2 (I_R2) can be calculated as:
\[ I_{R2} = \frac{E}{R2} \]

**Hint:** The total current supplied by the battery is the sum of the currents through each parallel resistor.

### Step 5: Determine the Total Current from the Battery
The total current (I_total) delivered by the battery is the sum of the currents through R and R2:
\[ I_{total} = I_R + I_{R2} = \frac{E}{R} + \frac{E}{R2} \]

**Hint:** If R2 is equal to R, the total current will be double that of the current through a single resistor.

### Step 6: Analyze the Power Dissipated by Resistor R
The power dissipated by resistor R (P_R) can be calculated using the formula:
\[ P_R = \frac{V^2}{R} = \frac{E^2}{R} \]
Since the voltage across R remains the same, the power dissipated does not change when R2 is added in parallel.

**Hint:** The power remains the same as long as the voltage across the resistor does not change.

### Conclusion
1. The potential across R remains constant (equal to E).
2. The current through R does not decrease; it remains the same as before.
3. The total current delivered by the battery increases.
4. The power dissipated by R remains the same.

### Final Answer
The correct statement is that the current delivery by the battery will increase.