So this is what I did for the first bit. Ignore that misplaced T1 at the right.
Then for the next bit
\(T_{1}=0.69C(R_{b})
30*10^{-3}=0.69(3.3*10^8)R_{b}
30*10^{-3}=2.277*10^{-6}R_{b}
\frac{30*10^{-3}}{2.277*10^{-6}}=R_{b}
R_{b}\sim 13k\)
Right for the last question I am really not sure how the mark scheme got \(3R_{b}=R_{a} \) , I really can't see how they derived that.
Then for the next bit
\(T_{1}=0.69C(R_{b})
30*10^{-3}=0.69(3.3*10^8)R_{b}
30*10^{-3}=2.277*10^{-6}R_{b}
\frac{30*10^{-3}}{2.277*10^{-6}}=R_{b}
R_{b}\sim 13k\)
Right for the last question I am really not sure how the mark scheme got \(3R_{b}=R_{a} \) , I really can't see how they derived that.
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