1) Wat is the total resistance between node A - B2) What is the total current between node A - B3) What is the voltage drop across R1, R2 ### Circuit Diagram ExplanationThis image depicts a simple electrical circuit with the following components and notations:1. **Voltage Source \( V_1 \):** - Represents a voltage supply of 9 volts, shown as a circle with a plus and minus sign. 2. **Resistors:** - **\( R_1 \):** A resistor with a resistance of 4 ohms. - **\( R_2 \):** A resistor with a resistance of 8 ohms. - **\( R_3 \):** A resistor with a resistance of 10 ohms.3. **Nodes:** - **Node A:** Located at the top left of the circuit, before the first resistor \( R_1 \). - **Node B:** Positioned at the bottom left of the circuit, after the last resistor \( R_3 \).These components are connected in series. The diagram is drawn on a grid background, providing a clear visual representation of the circuit's layout.

Resistance opposes the flow of current. KVL (Kirchhoff's Voltage Law) states that the algebraic sum of all voltage in a closed loop is zero. It is based on the conservation of energy. KCL (Kirchhoff's Current Law) states that the algebraic sum of current at node or junction is zero. It is based on the conservation of charge.1. If n number of resistor having resistance R1,R2.....................Rn is connected in series then the equivalent resistance Rs Rs=R1+R2+.....Rn If n number of resistor having resistance R1,R2.....................Rn is connected in parallel then the equivalent resistance Rp 1RP=1R1+1R2+.....1Rn The resistance in circuit are connected in series the total resistance through A-B is evaluated as below RA-B=4+8+10=22Ω Therefore required value of resistance is 22Ω 2. The current through terminal A-B is obtained by applying KVL in circuit Consider the current I is flowing in the circuit in clockwise direction -9+4I+8I+10I=022I=9I=922A≈0.41A Therefore required current value is I=0.41A 3. The voltage drop V1 is evaluated by using Ohm's V1=IR1=922×4≈1.64V The voltage drop V2 is evaluated by using Ohm's V2=IR2=922×8≈3.27V Therefore required values are V1=1.64V,V2=3.27V