Let's break this down step-by-step: ### 1. **Understanding the Problem** - **Goal:** Design a voltage divider using an LDR and a fixed resistor so that the output voltage is **2.5V or higher** when it is "light" (i.e., when the LDR resistance is **7 kΩ**). - **LDR:** At the desired light level, its resistance is **7 kΩ**. - **Voltage Divider:** We want to use a voltage divider to produce the desired voltage. - **Assume:** The supply voltage (**Vcc**) is **5V** (this is typical for such circuits, and unless otherwise stated, it's a reasonable assumption). --- ### 2. **Circuit Arrangement** A voltage divider consists of two resistors in series between Vcc and GND. The output is taken from the junction between them. We have two options: - **Option A:** LDR on top (connected to Vcc), fixed resistor at the bottom (connected to GND). - **Option B:** Fixed resistor on top (connected to Vcc), LDR at the bottom (connected to GND). #### Which arrangement gives higher voltage when it is light? - When it is light, the LDR resistance **decreases** (here, 7 kΩ). - If LDR is on top: as it gets lighter, the top resistance drops, so output voltage **drops**. - If LDR is at the bottom: as it gets lighter, the bottom resistance drops, so output voltage **rises**. **Therefore, we should put the LDR as the lower resistor (between output and ground), and the fixed resistor on top (between Vcc and output).** --- ### 3. **Voltage Divider Equation** The voltage at the divider output is: \[ V_{out} = V_{cc} \times \frac{R_{LDR}}{R_{fixed} + R_{LDR}} \] But in our arrangement, since the fixed resistor is on top and LDR on bottom (to GND): \[ V_{out} = V_{cc} \times \frac{R_{LDR}}{R_{fixed} + R_{LDR}} \] Given: - \( V_{cc} = 5V \) - \( R_{LDR} = 7\,k\Omega \) (at target brightness) - \( V_{out} \geq 2.5V \) - Need to find \( R_{fixed} \) values (or value). --- ### 4. **Setting up the Equation** We want: \[ V_{out} \geq 2.5 V \] \[ 5V \times \frac{7k}{R_{fixed} + 7k} \geq 2.5V \] Divide both sides by 5V: \[ \frac{7k}{R_{fixed} + 7k} \geq 0.5 \] Multiply both sides by \( R_{fixed} + 7k \): \[ 7k \geq 0.5(R_{fixed} + 7k) \] Expand: \[ 7k \geq 0.5 R_{fixed} + 3.5k \] Subtract \(3.5k\) from both sides: \[ 3.5k \geq 0.5 R_{fixed} \] Multiply both sides by 2: \[ 7k \geq R_{fixed} \] So, \[ R_{fixed} \leq 7k\Omega \] --- ### 5. **Final Answer** **To get 2.5 V or higher at the output when the LDR is 7 kΩ:** - Use a voltage divider with the fixed resistor on top (between +5V and output) and the LDR on the bottom (between output and GND). - The fixed resistor should be **7 kΩ or less**. **Common E12 series resistor values less than or equal to 7 kΩ are:** - 6.8 kΩ - 6.2 kΩ - 5.6 kΩ - ...etc. **So, you can use a 6.8 kΩ fixed resistor.** --- ### 6. **Check with 6.8 kΩ Resistor** \[ V_{out} = 5V \times \frac{7k}{6.8k + 7k} = 5V \times \frac{7}{13.8} \] \[ 7/13.8 \approx 0.507 \] \[ V_{out} = 5 \times 0.507 \approx 2.54V \] Which is **above 2.5V** as required. --- ## **Summary Table** | LDR (at light) | Fixed Resistor | Output Voltage | |----------------|-----------------|---------------| | 7 kΩ | 6.8 kΩ | ~2.54 V | | 7 kΩ | 6.2 kΩ | ~2.61 V | | 7 kΩ | 5.6 kΩ | ~2.71 V | --- ## **Final Answer** **Connect the LDR between output and ground, and a fixed resistor (6.8 kΩ or less) between +5V and output. At 7 kΩ LDR resistance, this will give 2.5 V or higher at the output.** **Recommended value:** **6.8 kΩ fixed resistor.** **Circuit:** ``` +5V | [6.8 kΩ Resistor] | |---- V_out (to sensor, etc.) | [LDR (7 kΩ at threshold)] | GND ```