Two resistors of resistances `200 k Omega` and `1M Omega` respectively form a potential divider with outer junctions maintained at potentials of + 3V and −15 V. Then, the potential at the junction between the resistors is-

Options

A. `+ 1 V`
B. `-0.6 V`
C. `0 V`
D. `-12 V`

Explanation

To solve the problem of finding the potential at the junction between two resistors in a potential divider, we can follow these steps:

### Step 1: Identify the resistances and potentials
We have two resistors:
- \( R_1 = 200 \, k\Omega = 200 \times 10^3 \, \Omega \)
- \( R_2 = 1 \, M\Omega = 1 \times 10^6 \, \Omega \)

The outer potentials are:
- \( V_1 = +3 \, V \)
- \( V_2 = -15 \, V \)

### Step 2: Calculate the equivalent resistance
Since the resistors are in series, the equivalent resistance \( R_{eq} \) is given by:
\[ R_{eq} = R_1 + R_2 = 200 \times 10^3 + 1 \times 10^6 = 1.2 \times 10^6 \, \Omega \]

### Step 3: Calculate the total voltage across the resistors
The total voltage \( V_{total} \) across the resistors is the difference between the two potentials:
\[ V_{total} = V_1 - V_2 = 3 - (-15) = 3 + 15 = 18 \, V \]

### Step 4: Calculate the current flowing through the circuit
Using Ohm's law, the current \( I \) flowing through the circuit can be calculated as:
\[ I = \frac{V_{total}}{R_{eq}} = \frac{18 \, V}{1.2 \times 10^6 \, \Omega} = 15 \times 10^{-6} \, A = 15 \, \mu A \]

### Step 5: Calculate the potential drop across \( R_1 \)
The potential drop \( V_{R1} \) across the resistor \( R_1 \) can be calculated using Ohm's law:
\[ V_{R1} = I \times R_1 = 15 \times 10^{-6} \times 200 \times 10^3 = 3 \, V \]

### Step 6: Calculate the potential at the junction
The potential at the junction \( V_x \) (between \( R_1 \) and \( R_2 \)) can be found by subtracting the potential drop across \( R_1 \) from the potential at the positive end:
\[ V_x = V_1 - V_{R1} = 3 \, V - 3 \, V = 0 \, V \]

### Final Answer
The potential at the junction between the resistors is:
\[ \boxed{0 \, V} \]