Physics - Current Electricity Question with Solution | TestHub

Assuming $V_B=0\text{ V}$, then $V_A=12\text{ V}$. A. Current in branch ACB is $I_{ACB}=\frac{12}{R_1+R_2}=\frac{12}{2+4}=2\text{ A}$. So $V_C=V_A-I_{ACB}{R}_1=12-2(2)=8\text{ V}$. B. Current in branch ADB is $I_{ADB}=\frac{12}{R_3+R_4}=\frac{12}{3+6}=\frac{4}{3}\text{ A}$. So $V_D=V_A-I_{ADB}{R}_3=12-(\frac{4}{3})(3)=8\text{ V}$. C. Since $V_C=V_D=8\text{ V}$, the potential difference $V_{CD}=0$, so no current flows through $R_5$. D. If $R_1=1\Omega$, $I_{ACB}^{\prime }=\frac{12}{1+4}=2.4\text{ A}$. $V_C^{\prime }=12-2.4(1)=9.6\text{ V}$. $V_D$ remains $8\text{ V}$. Thus $V_{CD}^{\prime }=9.6-8=1.6\text{ V}\ne 0$.