Determine the voltage drop across the resistor `R_1` in the circuit given below with `E = 65V, R_1 = 50 OmegaR_2 = 100 Omega, R_3 = 100 Omega, and R_4 300 Omega.

Explanation

`R_3 and R_4` are in series. The combined resistance is `400 Omega.`
Now `R_2 (-100Omega)and 400Omega` resistance are in parallel. The
combined resistance is `(100xx400)//500 = 80Omega.`
Total resistance is `R = 80 +50 = 130 Omega`
`I = 65//130 = 1//2A.`
So, `V = IR_1 = 1.//2 xx50 = 25V`