Example Question - voltage
Here are examples of questions we've helped users solve.
Understanding Ohm's Law and Its Graphical Representation
<p>Ohm's Law states that the current (I) through a conductor between two points is directly proportional to the voltage (V) across the two points, and inversely proportional to the resistance (R) of the conductor. The mathematical equation representing Ohm's Law is \( V = IR \).</p> <p>To graphically represent Ohm's Law, one would typically plot voltage (V) on the vertical axis and current (I) on the horizontal axis. For a resistor with a constant resistance, the graph will be a straight line passing through the origin (0,0). This line indicates that as the voltage increases, the current through the resistor increases proportionally, and each point on the line stays consistent with Ohm's Law \( V = IR \). The slope of this line (rise over run) is equal to the resistance (R).</p>
Identifying Electrical Laws by Formula
<p>The formula \( I = \frac{V}{R} \) is known as Ohm's Law, which states that the current \( I \) through a conductor between two points is directly proportional to the voltage \( V \) across the two points and inversely proportional to the resistance \( R \) between them. Therefore, the correct answer is:</p> <p>\( \text{Hukum Ohm} \)</p>
Calculating Thevenin Equivalent Voltage and Resistance
To solve for the Thevenin equivalent voltage (V_th) and Thevenin equivalent resistance (R_th) for the given circuit, we will follow these steps: 1. Remove the load from the original circuit to isolate the part that will be represented by the Thevenin equivalent. In this case, the "LOAD" is the part we exclude to determine V_th and R_th. 2. Calculate the Thevenin voltage (V_th), which is the open-circuit voltage at the terminals where the load was connected. Since there is no current flowing through the 4 Ω resistor (because it is in series with the load and the load is removed), the voltage across the 4 Ω resistor is 0 V. Thus, the Thevenin voltage V_th is the same as the voltage of the voltage source, which is -10 V. 3. Calculate the Thevenin resistance (R_th) from the perspective of the load terminals with all independent voltage sources replaced by a short circuit. In this circuit, replacing the voltage source with a short circuit makes the 8 Ω and 4 Ω resistors in parallel. Their combined resistance, R_th, can be calculated using the formula for parallel resistors: \[ \frac{1}{R_{th}} = \frac{1}{R_1} + \frac{1}{R_2} \] \[ \frac{1}{R_{th}} = \frac{1}{8 \Omega} + \frac{1}{4 \Omega} \] \[ \frac{1}{R_{th}} = \frac{1 + 2}{8 \Omega} \] \[ \frac{1}{R_{th}} = \frac{3}{8 \Omega} \] \[ R_{th} = \frac{8}{3} \Omega \] So, the Thevenin equivalent resistance is \(\frac{8}{3} \Omega\) or approximately \(2.67 \Omega\). To summarize: Thevenin voltage (V_th) = -10 V Thevenin resistance (R_th) ≈ 2.67 Ω
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